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All questions of Probability Class 11 for JEE Exam
One card is drawn from a pack of cards, each of the 52 cards being equally likely to be drawn. The probability that the card drawn is red and a queen is:- a)1/26
- b)1/2
- c)2
- d)1/4
Correct answer is option 'A'. Can you explain this answer?
One card is drawn from a pack of cards, each of the 52 cards being equally likely to be drawn. The probability that the card drawn is red and a queen is:
a)
1/26
b)
1/2
c)
2
d)
1/4
|
Gaurav Kumar answered |
The cards contains 4 Queen from which 2 are black and 2 are red
we need to find the probability that the card drawn is red and a queen is: 2/52
= 1/26
we need to find the probability that the card drawn is red and a queen is: 2/52
= 1/26
Probability of getting a number between 1 and 100, which is divisible by 1 and itself only, is- a)25/99
- b)1/4
- c)25/98
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Probability of getting a number between 1 and 100, which is divisible by 1 and itself only, is
a)
25/99
b)
1/4
c)
25/98
d)
none of these
|
Nandini Iyer answered |
The prime numbers between 1 and 100 are :2, 3, 5, 7, 11,
A box contains 15 red marbles, 15 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that atleast one will be green?- a)
- b)
- c) zero
- d)1
Correct answer is option 'A'. Can you explain this answer?
A box contains 15 red marbles, 15 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that atleast one will be green?
a)
b)
c)
zero
d)
1
|
Shashwat Singh answered |
It's like doing the question orally... (don't get into maths... as I was also stuck in the middle) but by common sense (which is very uncommon) it can be said as 1-probablity of not getting any green in 5 balls... in this case we can see the answer aproaches to 1 which can be seen in option a) .... hope it helps... but with maths a can see a nightmare coming...
The probability that a man will live for 10 more years is 1/4 and that his wife will live 10 more years is 1/3. The probability that neither will be alive in 10 years is- a)5/12
- b)1/2
- c)11/12
- d)7/12
Correct answer is option 'B'. Can you explain this answer?
The probability that a man will live for 10 more years is 1/4 and that his wife will live 10 more years is 1/3. The probability that neither will be alive in 10 years is
a)
5/12
b)
1/2
c)
11/12
d)
7/12
|
Mayank Dasgupta answered |
Given probabilities:
- Probability that the man will live for 10 more years = 1/4
- Probability that his wife will live for 10 more years = 1/3
To find: Probability that neither will be alive in 10 years
Solution:
Let A be the event that the man will be alive in 10 years, and B be the event that his wife will be alive in 10 years. Then, the probability that neither will be alive in 10 years is the probability of the complement of the union of A and B, i.e., P((A ∪ B)').
Using the formula for the probability of the union of two events, we have:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
where P(A ∩ B) is the probability that both the man and his wife will be alive in 10 years.
Substituting the given probabilities, we get:
P(A ∪ B) = 1/4 + 1/3 - P(A ∩ B)
Simplifying, we get:
P(A ∩ B) = 7/12 - P((A ∪ B)')
Now, we know that the probability of the man being alive in 10 years is 1/4, which means the probability of him not being alive in 10 years is 3/4. Similarly, the probability of his wife not being alive in 10 years is 2/3.
Using the product rule of probability, we can find the probability of both of them not being alive in 10 years:
P((A ∪ B)') = P(A' ∩ B') = P(A') × P(B') = (3/4) × (2/3) = 1/2
Substituting this value in the above equation, we get:
P(A ∩ B) = 7/12 - 1/2 = 1/12
Therefore, the probability that neither the man nor his wife will be alive in 10 years is:
P((A ∪ B)') = 1 - P(A ∪ B) = 1 - (1/4 + 1/3 - 1/12) = 1/2
Hence, the correct option is (B).
- Probability that the man will live for 10 more years = 1/4
- Probability that his wife will live for 10 more years = 1/3
To find: Probability that neither will be alive in 10 years
Solution:
Let A be the event that the man will be alive in 10 years, and B be the event that his wife will be alive in 10 years. Then, the probability that neither will be alive in 10 years is the probability of the complement of the union of A and B, i.e., P((A ∪ B)').
Using the formula for the probability of the union of two events, we have:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
where P(A ∩ B) is the probability that both the man and his wife will be alive in 10 years.
Substituting the given probabilities, we get:
P(A ∪ B) = 1/4 + 1/3 - P(A ∩ B)
Simplifying, we get:
P(A ∩ B) = 7/12 - P((A ∪ B)')
Now, we know that the probability of the man being alive in 10 years is 1/4, which means the probability of him not being alive in 10 years is 3/4. Similarly, the probability of his wife not being alive in 10 years is 2/3.
Using the product rule of probability, we can find the probability of both of them not being alive in 10 years:
P((A ∪ B)') = P(A' ∩ B') = P(A') × P(B') = (3/4) × (2/3) = 1/2
Substituting this value in the above equation, we get:
P(A ∩ B) = 7/12 - 1/2 = 1/12
Therefore, the probability that neither the man nor his wife will be alive in 10 years is:
P((A ∪ B)') = 1 - P(A ∪ B) = 1 - (1/4 + 1/3 - 1/12) = 1/2
Hence, the correct option is (B).
The sample space of an experiment is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} and A = {TTT}, B = {HTT, THT, TTH}, C = {HHT, HTH, THH, HHH},
then A, B and C form a set of?- a)Mutually exclusive but not exhaustive events
- b)Exhaustive but not mutually exclusive events
- c)Mutually exclusive and exhaustive events
- d)Neither mutually exclusive nor exhaustive events
Correct answer is option 'C'. Can you explain this answer?
The sample space of an experiment is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} and A = {TTT}, B = {HTT, THT, TTH}, C = {HHT, HTH, THH, HHH},
then A, B and C form a set of?
then A, B and C form a set of?
a)
Mutually exclusive but not exhaustive events
b)
Exhaustive but not mutually exclusive events
c)
Mutually exclusive and exhaustive events
d)
Neither mutually exclusive nor exhaustive events
|
Suresh Reddy answered |
Above information has no common element in three of them and U of three sets forms sample set . Therefore option C is correct.
A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability that the lost card was a diamond:- a)144/169
- b)11/50
- c)39/50
- d)11/24
Correct answer is option 'B'. Can you explain this answer?
A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability that the lost card was a diamond:
a)
144/169
b)
11/50
c)
39/50
d)
11/24
|
Defence Exams answered |
Let D – Lost card is diamond
D – Lost card is not diamond.
When one diamond card is lost, the number of diamond cards are 12 out of 51.
No. of ways of choosing 2 from 12 cards 12C2 .
Similarity, choosing 2 diamond cards out of 51 are 51C2.
probability of getting two cards, when one diamond is lost
When the lost and is not diamond, number of diamond cards are 13 out of 51.
No. of ways of choosing 2 cards when one card is lost which is not diamond =
D – Lost card is not diamond.
When one diamond card is lost, the number of diamond cards are 12 out of 51.
No. of ways of choosing 2 from 12 cards 12C2 .
Similarity, choosing 2 diamond cards out of 51 are 51C2.
probability of getting two cards, when one diamond is lostWhen the lost and is not diamond, number of diamond cards are 13 out of 51.
No. of ways of choosing 2 cards when one card is lost which is not diamond =
If 2/7 is the probability of an event, then the probability of the event ‘not A’ is:- a)5/7
- b)6/7
- c)2/7
- d)3/7
Correct answer is option 'A'. Can you explain this answer?
If 2/7 is the probability of an event, then the probability of the event ‘not A’ is:
a)
5/7
b)
6/7
c)
2/7
d)
3/7
|
|
Geetika Shah answered |
If 2/7 is probability of event A then the probability of an event not A is 1-(2/7) i.e., probability of not A is 5/7
From a group of 5 boys and 3 girls, 3 persons are chosen at random. Then the probability that there are more girls than boys is:- a)15/28
- b)15/56
- c)1
- d)2/7
Correct answer is option 'D'. Can you explain this answer?
From a group of 5 boys and 3 girls, 3 persons are chosen at random. Then the probability that there are more girls than boys is:
a)
15/28
b)
15/56
c)
1
d)
2/7
|
Tejas Verma answered |
There are 3 girls and 5 boys
∴ Total number of boys and girls = 8
n(s) = Total number of ways of choosing 3 persons
Let E be the event that there are more girls than boys.
⇒ (2 girls & 1 boy) or (3 girls & 0 boy)
∴ Total number of boys and girls = 8
n(s) = Total number of ways of choosing 3 persons
Let E be the event that there are more girls than boys.
⇒ (2 girls & 1 boy) or (3 girls & 0 boy)
Probability that A speaks truth is 4/5. A coin is tossed, a reports that a head appears. The probability that actually there was head is- a)1
- b)1/2
- c)1/5
- d)4/5
Correct answer is option 'D'. Can you explain this answer?
Probability that A speaks truth is 4/5. A coin is tossed, a reports that a head appears. The probability that actually there was head is
a)
1
b)
1/2
c)
1/5
d)
4/5
|
|
Naina Sharma answered |
Let E1 : Head appears
E2 : Tail appears
A : A reports that head appears
∴ Rwquired probability =
By Bayes’ Theorem
E2 : Tail appears
A : A reports that head appears
∴ Rwquired probability =
By Bayes’ Theorem
The events when we have no reason to believe that one is more likely to occur than the other is called:- a)Equally likely events
- b)Independent events
- c)Dependent event
- d)Not equally likely events
Correct answer is option 'A'. Can you explain this answer?
The events when we have no reason to believe that one is more likely to occur than the other is called:
a)
Equally likely events
b)
Independent events
c)
Dependent event
d)
Not equally likely events
|
Vikas Kapoor answered |
Equally Likely Events Events which have the same chance of occurring Probability. Chance that an event will occur. Theoretically for equally likely events, it is the number of ways an event can occur divided by number of outcomes in the sample space.
A and B throw a coin alternately till one of them gets a ‘head’ & wins the game. If A has the first threw, then the probabilities of winning for A and B are respectively.- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A and B throw a coin alternately till one of them gets a ‘head’ & wins the game. If A has the first threw, then the probabilities of winning for A and B are respectively.
a)
b)
c)
d)
|
Praveen Kumar answered |
Let S be success &F be failure and let p = probability of getting head = 1/2
q = 1/2
A starts the game then
P (A wins) = P (A wins in first throw) + P (A wins in third throw) + P (A wins in fifth throw) +......... ∞
= P(S) + P(FFS) + P(FFFFS) +......... ∞
= p + q.q.p + q.q.q.q.p +.........
This is an infinite GP and hence its sum is
q = 1/2
A starts the game then
P (A wins) = P (A wins in first throw) + P (A wins in third throw) + P (A wins in fifth throw) +......... ∞
= P(S) + P(FFS) + P(FFFFS) +......... ∞
= p + q.q.p + q.q.q.q.p +.........
This is an infinite GP and hence its sum is
is equal to
In a simultaneous toss of two coins, the probability of getting no tail is:- a)1/4
- b)1/2
- c)2
- d)0.1
Correct answer is option 'A'. Can you explain this answer?
In a simultaneous toss of two coins, the probability of getting no tail is:
a)
1/4
b)
1/2
c)
2
d)
0.1
|
Om Desai answered |
Sample space = {HH, HT, TH, TT}
n(SS) = 4
No tail = {HH}
n(No tail) = 1
P(No tail) = n(No tail) / n(SS
= 1/4
n(SS) = 4
No tail = {HH}
n(No tail) = 1
P(No tail) = n(No tail) / n(SS
= 1/4
A letter is known to have come either from TATANAGAR or from CALCUTTA. On the envelope just two consecutive letters TA are visible. What is the probability that the letters came from TATANAGAR ?- a)7/11
- b)4/11
- c)24/29
- d)1/11
Correct answer is option 'A'. Can you explain this answer?
A letter is known to have come either from TATANAGAR or from CALCUTTA. On the envelope just two consecutive letters TA are visible. What is the probability that the letters came from TATANAGAR ?
a)
7/11
b)
4/11
c)
24/29
d)
1/11
|
Amrita Sharma answered |
Ans.
Option (a)
Let E1 denote the event that the letter came from TATA NAGAR and E2 the event that the letter came from CALCUTTA.
Let A denote the event that the two consecutive alphabets visible on the envelope are TA.
Since, the letters have come either from Calcutta or Tata nagar, Therefore,
P (E1) = � and P (E2) = �
If E1 has occurred, then it means that the letter came from TATA NAGAR. In the word TATA NAGAR there are 8 consecutive alphabets. i.e., {TA, AT, TA, AN, NA, AG, GA and AR} and TA occurs two time.
Therefore, P (A|E1) = 2/8=1/4
If E2 has occurred, then it means that the letter came from Calcutta. In the word CALCUTTA there are 7 consecutive alphabets. i.e., {CA, AL, LC, CU, UT, TT and TA} and TA occurs only one time.
Therefore, P (A|E2) = 1/7
P (A|E1) = 2/8, P (A|E2) = 1/7. Therefore, by Bayes theorem we have,
Hence, probability that the letter has come from CALCUTTA is 4/11.
And,
Hence, probability that the letter has come from TATA NAGAR is 7/11.
Bag I contains 3 red and 4 black balls and bag II contains 4 red and 5 blacks’ balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red. Find the probability that the transferred ball is black- a)16/31
- b)15/31
- c)3/8
- d)5/8
Correct answer is option 'A'. Can you explain this answer?
Bag I contains 3 red and 4 black balls and bag II contains 4 red and 5 blacks’ balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red. Find the probability that the transferred ball is black
a)
16/31
b)
15/31
c)
3/8
d)
5/8
|
Anjali Sharma answered |
Step 1:
Given Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball drawn is red in color.
Let E1 be the event that a red ball is transferred from Bag I to Bag II, and E2 be the event that a black ball is transferred from Bag I to Bag II. Let A be the event that the ball drawn is red.
P(E1)=3/3+4=3/7
P(E2)=4/4+3=4/7
Step 2:
P(A/E1)=5/10=1/2
P(A/E2)=4/10=2/5
P(E2/E) = P(E/E2).P(E2)/∑4i1(P(EEi) . P(Ei))
⇒ 4/7x2/5 / 3/7x1/2 + 4/7x2/5 = 16/31
If an event has more than one sample point, it is called a ………- a)Exhaustive event
- b)Compound event
- c)Simple event
- d)Mutually exclusive event
Correct answer is option 'B'. Can you explain this answer?
If an event has more than one sample point, it is called a ………
a)
Exhaustive event
b)
Compound event
c)
Simple event
d)
Mutually exclusive event
|
Infinity Academy answered |
Compound Event If an event has more than one sample point it is called a compound event, for example, S = {HH, HT} is a compound event.
Two students Amit and Priyanka appeared in an examination. The probability that Amit will qualify the examination is 0.06 and that Priyanka will qualify the examination is 0.12. The probability that both will qualify the examination is 0.02. Then, the probability that both Amit and Priyanka will not qualify the examination is:- a)0.91
- b)0.12
- c)0.84
- d)0.45
Correct answer is option 'C'. Can you explain this answer?
Two students Amit and Priyanka appeared in an examination. The probability that Amit will qualify the examination is 0.06 and that Priyanka will qualify the examination is 0.12. The probability that both will qualify the examination is 0.02. Then, the probability that both Amit and Priyanka will not qualify the examination is:
a)
0.91
b)
0.12
c)
0.84
d)
0.45
|
|
Naina Sharma answered |
Probability that Amit will qualify the exam(E) = 0.06
Probability that Priyanka will qualify the exam(F) = 0.12
Probability that both will qualify the exam P(E⋂F) = 0.02
P(EUF) = P(E) + P(F) - P(EUF)
= 0.06 + 0.12 - 0.02 = 0.16
P(E’⋂F’) = 1 - P(EUF)
⇒ 1 - 0.16
= 0.84
Probability that Priyanka will qualify the exam(F) = 0.12
Probability that both will qualify the exam P(E⋂F) = 0.02
P(EUF) = P(E) + P(F) - P(EUF)
= 0.06 + 0.12 - 0.02 = 0.16
P(E’⋂F’) = 1 - P(EUF)
⇒ 1 - 0.16
= 0.84
When the sets A and B are two events associated with a sample space, then "A ∩ B " is the event common to
- a)Either A or B or both
- b)Only one of the two, A or B
- c)Both A and B
- d)Neither A nor B
Correct answer is option 'C'. Can you explain this answer?
When the sets A and B are two events associated with a sample space, then "A ∩ B " is the event common to
a)
Either A or B or both
b)
Only one of the two, A or B
c)
Both A and B
d)
Neither A nor B
|
Akanksha Reddy answered |
Understanding Events A and B
In probability theory, events are subsets of a sample space. When we talk about two events A and B, we can analyze their relationship through various operations, one of which is the intersection denoted by "A ∩ B".
What is A ∩ B?
- The intersection "A ∩ B" refers to the set of outcomes that are common to both events A and B.
- This means that for an outcome to be part of "A ∩ B", it must satisfy the conditions of both events simultaneously.
Why is the Correct Answer C?
- The correct answer is option 'C', which states that "A ∩ B" represents outcomes that occur in both events A and B.
- In simpler terms, if you were to visualize events A and B in a Venn diagram, "A ∩ B" would be the overlapping area where both circles intersect.
Clarifying the Other Options
- a) "Either A or B or both" refers to the union of events A and B (A ∪ B), which includes all outcomes in A, B, or both.
- b) "Only one of the two, A or B" suggests a scenario where outcomes belong exclusively to one event, which is not what "A ∩ B" denotes.
- d) "Neither A nor B" indicates outcomes that do not belong to either event, which again does not describe the intersection.
Conclusion
In summary, "A ∩ B" specifically captures the essence of both events occurring together. Therefore, the assertion that "A ∩ B" is the event common to both A and B is accurately represented by option 'C'. Understanding these relationships is fundamental in probability and helps in solving complex problems in the JEE syllabus.
In probability theory, events are subsets of a sample space. When we talk about two events A and B, we can analyze their relationship through various operations, one of which is the intersection denoted by "A ∩ B".
What is A ∩ B?
- The intersection "A ∩ B" refers to the set of outcomes that are common to both events A and B.
- This means that for an outcome to be part of "A ∩ B", it must satisfy the conditions of both events simultaneously.
Why is the Correct Answer C?
- The correct answer is option 'C', which states that "A ∩ B" represents outcomes that occur in both events A and B.
- In simpler terms, if you were to visualize events A and B in a Venn diagram, "A ∩ B" would be the overlapping area where both circles intersect.
Clarifying the Other Options
- a) "Either A or B or both" refers to the union of events A and B (A ∪ B), which includes all outcomes in A, B, or both.
- b) "Only one of the two, A or B" suggests a scenario where outcomes belong exclusively to one event, which is not what "A ∩ B" denotes.
- d) "Neither A nor B" indicates outcomes that do not belong to either event, which again does not describe the intersection.
Conclusion
In summary, "A ∩ B" specifically captures the essence of both events occurring together. Therefore, the assertion that "A ∩ B" is the event common to both A and B is accurately represented by option 'C'. Understanding these relationships is fundamental in probability and helps in solving complex problems in the JEE syllabus.
Five marbles are drawn from a bag which contains 6 blue marbles and 7 green marbles. Then, the probability that 3 will be blue and 2 green is:- a)213/429
- b)140/429
- c)117/429
- d)167/429
Correct answer is option 'B'. Can you explain this answer?
Five marbles are drawn from a bag which contains 6 blue marbles and 7 green marbles. Then, the probability that 3 will be blue and 2 green is:
a)
213/429
b)
140/429
c)
117/429
d)
167/429
|
|
Naina Sharma answered |
Blue = 6
Green = 7
Number of ways 3 blue marbles can be drawn = 6C3
number of ways 2 green marbles can be drawn = 7C2
total ways of drawing 5 marbles = 13C5
=> probability = (6C3 * 7C2)/13C5
= 140/429
Green = 7
Number of ways 3 blue marbles can be drawn = 6C3
number of ways 2 green marbles can be drawn = 7C2
total ways of drawing 5 marbles = 13C5
=> probability = (6C3 * 7C2)/13C5
= 140/429
Six boys and six girls sit in a row at random. Then the probability that six girls sit together is- a)1/132
- b)1/32
- c)1/360
- d)1/36
Correct answer is option 'A'. Can you explain this answer?
Six boys and six girls sit in a row at random. Then the probability that six girls sit together is
a)
1/132
b)
1/32
c)
1/360
d)
1/36
|
Nishanth Verma answered |
Let n (s) = total number of ways of seating (permutations) 6 boys and 6 girls
i.e. 12 people = 12P12 = 12!
Let E: six girls sit together.
Then they become one unit
∴ No. of units is 6 boys + 1 unit of girls = 7 units
∴ n (E) = No. of ways in which six girls can sit together is = 7!, 6!
(Here 6! is the number of ways in which the six girls can sit among themselves)
i.e. 12 people = 12P12 = 12!
Let E: six girls sit together.
Then they become one unit
∴ No. of units is 6 boys + 1 unit of girls = 7 units
∴ n (E) = No. of ways in which six girls can sit together is = 7!, 6!
(Here 6! is the number of ways in which the six girls can sit among themselves)
= 2/3, then 
8 coins are tossed at a time. The probability of getting 6 heads up is- a)57/64
- b)7/64
- c)37/256
- d)229/256
Correct answer is option 'C'. Can you explain this answer?
8 coins are tossed at a time. The probability of getting 6 heads up is
a)
57/64
b)
7/64
c)
37/256
d)
229/256
|
|
Koyna Jatwar answered |
The required probability =8C6(1/2)^6.(1/2)^2+8C7(1/2)^7(1/2)+8C8(1/2)^8. =1/256(28+8+1)=37/256
The letters of the word ‘ASSASSIN ‘ are written at random in a row. The chance that all the similar letters occur together is- a)1/35
- b)1/14
- c)34/35
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The letters of the word ‘ASSASSIN ‘ are written at random in a row. The chance that all the similar letters occur together is
a)
1/35
b)
1/14
c)
34/35
d)
none of these
|
|
Anjali Sharma answered |
Total number of ways to arrange "ASSASSIN" is 8!/(4!.2!) first we fix the position.
A bag contains 5 brown and 4 white socks. A man pulls out 2 socks. The probability that they are of the same colour is:- a)5/108
- b)4/9
- c)2/9
- d)5/9
Correct answer is option 'B'. Can you explain this answer?
A bag contains 5 brown and 4 white socks. A man pulls out 2 socks. The probability that they are of the same colour is:
a)
5/108
b)
4/9
c)
2/9
d)
5/9
|
Nilotpal Ghoshal answered |
Problem Analysis
The problem states that there are 5 brown socks and 4 white socks in a bag. A man pulls out 2 socks randomly from the bag. We need to find the probability that both socks are of the same color.
Possible Outcomes
In order to find the probability, we first need to determine the total number of possible outcomes.
Since the man pulls out 2 socks, the total number of possible outcomes can be calculated using the combination formula:
Total outcomes = nCr = (n!) / (r!(n-r)!), where n is the total number of socks and r is the number of socks being pulled out.
In this case, n = 9 (5 brown + 4 white socks) and r = 2 (2 socks being pulled out). Therefore, the total number of possible outcomes is:
Total outcomes = 9C2 = (9!) / (2!(9-2)!) = 36
Favorable Outcomes
Next, we need to determine the number of favorable outcomes, i.e., the number of outcomes where both socks are of the same color.
Case 1: Both socks are brown:
In this case, there are 5 brown socks to choose from for the first sock and 4 brown socks remaining to choose from for the second sock. Therefore, the number of outcomes where both socks are brown is:
Number of outcomes = 5C1 * 4C1 = (5!) / (1!(5-1)!) * (4!) / (1!(4-1)!) = 20
Case 2: Both socks are white:
Similarly, the number of outcomes where both socks are white is:
Number of outcomes = 4C1 * 3C1 = (4!) / (1!(4-1)!) * (3!) / (1!(3-1)!) = 12
Probability Calculation
Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = (Number of favorable outcomes) / (Total number of possible outcomes)
Probability = (20 + 12) / 36
Probability = 32 / 36
Probability = 4 / 9
Therefore, the probability that the man pulls out 2 socks of the same color is 4/9, which corresponds to option B.
The letters of the word ‘MALENKOV‘are arranged in all possible ways. The chance that there are exactly four letters between M and E is- a)3/28
- b)3/14
- c)1/14
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The letters of the word ‘MALENKOV‘are arranged in all possible ways. The chance that there are exactly four letters between M and E is
a)
3/28
b)
3/14
c)
1/14
d)
none of these
|
Shilpa Rane answered |
"algorithm" can be rearranged in 9! (362,880) different ways.
The probability that a card drawn at random from a pack of 52 cards is a king or a heart is- a)16/52
- b)1/13
- c)1/52
- d)1/4
Correct answer is option 'A'. Can you explain this answer?
The probability that a card drawn at random from a pack of 52 cards is a king or a heart is
a)
16/52
b)
1/13
c)
1/52
d)
1/4
|
Er Aarif answered |
The number of kings in a deck are 4
Number of hearts in the deck are 13, including the king of hearts
Probability of getting either a king or a heart is, P = 4+(13–1) / 52
P = 16/52
Number of hearts in the deck are 13, including the king of hearts
Probability of getting either a king or a heart is, P = 4+(13–1) / 52
P = 16/52
From a pack of 52 cards, the cards are drawn one by one till an ace appears. The chance that an ace does not come up in first 26 cards is- a)109/153
- b)23/27
- c)46/153
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
From a pack of 52 cards, the cards are drawn one by one till an ace appears. The chance that an ace does not come up in first 26 cards is
a)
109/153
b)
23/27
c)
46/153
d)
none of these
|
Madhavan Chatterjee answered |
There are 4 aces in a pack. So, the probability of drawing an ace = 4/52
= 1/13
and the probability of drawing a card other than ace = 12/13.
Hence, the probability that ace doesn't appear in first 26 draws = (12/13)26
= 1/13
and the probability of drawing a card other than ace = 12/13.
Hence, the probability that ace doesn't appear in first 26 draws = (12/13)26
What is the sample space for an experiment when a coin is tossed and then a dice is thrown?- a){1H,2H3H,4H,5H,6H,T1,T2,T3,T4,T5,T6}
- b){H1,T1,H6,T6}
- c){H1,T1,H2,T2,H3,T3,H4,T4,H5,T5,H6,T6}
- d){HT,TH,TT,HH,1,2,3,4,5,6}
Correct answer is option 'C'. Can you explain this answer?
What is the sample space for an experiment when a coin is tossed and then a dice is thrown?
a)
{1H,2H3H,4H,5H,6H,T1,T2,T3,T4,T5,T6}
b)
{H1,T1,H6,T6}
c)
{H1,T1,H2,T2,H3,T3,H4,T4,H5,T5,H6,T6}
d)
{HT,TH,TT,HH,1,2,3,4,5,6}
|
|
Nandini Iyer answered |
When a coin is tossed.Head or Tail may occur.
Whereby when a die is thrown the numbers from 1 to 6 may occur.
H represents Head.
T represents Tail.
(1-6) sides of the die.
If the probabilities that A&B will die within a year are x & y respectively, then the probability that exactly one of them will be alive at the end of the year is- a)x + y – 2xy
- b)xy
- c)(1–x)(1–y)
- d)1–xy
Correct answer is option 'A'. Can you explain this answer?
If the probabilities that A&B will die within a year are x & y respectively, then the probability that exactly one of them will be alive at the end of the year is
a)
x + y – 2xy
b)
xy
c)
(1–x)(1–y)
d)
1–xy
|
Milan Roy answered |
P (A will die within a year) = x
P (B will die within a year) = y
∴ P (Exactly one of them will be alive at the end of the year)
= P (A is alive & B is dead)+ P (A dies B is alive)
(As these are independent events)
P (B will die within a year) = y
∴ P (Exactly one of them will be alive at the end of the year)
= P (A is alive & B is dead)+ P (A dies B is alive)
(As these are independent events)
A fair coin is tossed a fixed number of times. If the probability of getting 4 heads is equal to the probability of getting 7 heads, then the probability of getting 2 heads is- a)55/2048
- b)3/4096
- c)1/1024
- d)3/1024
Correct answer is option 'A'. Can you explain this answer?
A fair coin is tossed a fixed number of times. If the probability of getting 4 heads is equal to the probability of getting 7 heads, then the probability of getting 2 heads is
a)
55/2048
b)
3/4096
c)
1/1024
d)
3/1024
|
Sanchita Khanna answered |
Given, probability of getting 4 heads = probability of getting 7 heads
Let the number of tosses be n. Then,
Probability of getting 4 heads = Probability of getting (n-4) tails = (1/2)^n * nC4
Probability of getting 7 heads = Probability of getting (n-7) tails = (1/2)^n * nC7
Equating the two probabilities, we get
nC4 = nC7/35
Solving this equation, we get n = 11
So, the number of tosses is 11.
Now, we need to find the probability of getting 2 heads.
Probability of getting 2 heads = Probability of getting (11-2) tails = (1/2)^11 * nC2
= (1/2)^11 * 55
= 55/2048
Therefore, the correct option is A) 55/2048.
Let the number of tosses be n. Then,
Probability of getting 4 heads = Probability of getting (n-4) tails = (1/2)^n * nC4
Probability of getting 7 heads = Probability of getting (n-7) tails = (1/2)^n * nC7
Equating the two probabilities, we get
nC4 = nC7/35
Solving this equation, we get n = 11
So, the number of tosses is 11.
Now, we need to find the probability of getting 2 heads.
Probability of getting 2 heads = Probability of getting (11-2) tails = (1/2)^11 * nC2
= (1/2)^11 * 55
= 55/2048
Therefore, the correct option is A) 55/2048.
An experiment involves rolling a pair of dice and recording the number that comes up. Suppose,
A: the sum is greater than 8.
B: 2 occurs on either die. Then A and B are ……. events.- a)Random experiment
- b)Mutually exclusive events
- c)Exhaustive events
- d)Favorable events
Correct answer is option 'B'. Can you explain this answer?
An experiment involves rolling a pair of dice and recording the number that comes up. Suppose,
A: the sum is greater than 8.
B: 2 occurs on either die. Then A and B are ……. events.
A: the sum is greater than 8.
B: 2 occurs on either die. Then A and B are ……. events.
a)
Random experiment
b)
Mutually exclusive events
c)
Exhaustive events
d)
Favorable events
|
Advait Ghoshal answered |
A pair of dice is rolled
Therefore the sample space S will be
Pairs which are mutually exclusive :
A∩B=ϕ⇒ A and B are mutually exclusive.
B∩C=ϕ⇒ B and C are mutually exclusive.
A∩C=ϕ⇒ They are not mutually exclusive.
∴A∩B and B∩C are mutually exclusive.
A drawer contains 5 black socks and 4 blue socks well mixed. A person searches the drawer and pulls out 2 socks at random. The probability that they match is- a)41/81
- b)5/8
- c)4/9
- d)5/9
Correct answer is option 'C'. Can you explain this answer?
A drawer contains 5 black socks and 4 blue socks well mixed. A person searches the drawer and pulls out 2 socks at random. The probability that they match is
a)
41/81
b)
5/8
c)
4/9
d)
5/9
|
|
Jaya Deshpande answered |
Out of 9 socks, 2 can be drawn in 9C2 ways.
Therefore, the total number of cases is 9C2.
Two socks drawn from the drawer will match if either both are black or both are blue.
Therefore, favorable number of cases is 5C2+ 4C2.
Hence, the required probability is
(5C2+ 4C2.)/ 9C2
= 4/9
Three of the 6 vertices of a regular hexagon are chosen at random.The probability that the triangle with these vertices is equilateral is- a)1/10
- b)1/20
- c)1/5
- d)1/2
Correct answer is option 'A'. Can you explain this answer?
Three of the 6 vertices of a regular hexagon are chosen at random.The probability that the triangle with these vertices is equilateral is
a)
1/10
b)
1/20
c)
1/5
d)
1/2
|
|
Kurakula Suryateja answered |
Rules of Triangle …..
Total possible cases = 6 C 3
Favourable cases =2
So the probability is = 2/6 C 3 =1/10.
Because there can be 2 22 equilateral triangles possible and total number of triangles possible would be (6/3) the answer is 1/10.
Total possible cases = 6 C 3
Favourable cases =2
So the probability is = 2/6 C 3 =1/10.
Because there can be 2 22 equilateral triangles possible and total number of triangles possible would be (6/3) the answer is 1/10.
Two integers are selected at random from integers 1 to 11. If the sum is even, find the probability that both the numbers are odd.- a)3/11
- b)1
- c)3/5
- d)2/5
Correct answer is option 'C'. Can you explain this answer?
Two integers are selected at random from integers 1 to 11. If the sum is even, find the probability that both the numbers are odd.
a)
3/11
b)
1
c)
3/5
d)
2/5
|
Bhavya Joshi answered |
Out of integers from 1 to 11, there are 5 even and 6 odd integers.
Let A – both the numbers chosen are odd
B –sum of the numbers chosen is even.
Let A – both the numbers chosen are odd
B –sum of the numbers chosen is even.
A bag contains 5 white, 7 red and 4 black balls. Four balls are drawn one by one with replacement. The chance that atleast two balls are black is- a)243/256
- b)67/256
- c)54/256
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
A bag contains 5 white, 7 red and 4 black balls. Four balls are drawn one by one with replacement. The chance that atleast two balls are black is
a)
243/256
b)
67/256
c)
54/256
d)
none of these
|
Riya Banerjee answered |
Let A, B, C, D denote the events of not getting a white ball in first, second, third and fourth draw respectively.
Since the balls are drawn with replacement, therefore, A, B, C, D are independent events such that P (A) = P (B) = P (C) = P (D).
Since out of 16 balls, 11 are not white, therefore, P (A) = 11/16
∴ Required probability = P (A) . P (B) . P (C) . P (D)
=> (11/16) x (11/16) x (11/16) x (11/16) = (11/16)^4.
A man alternately tosses a coin and throws a dice beginning with the coin. The probability that he gets a head in the coin before he gets a 5 or 6 in the dice is- a)3/4
- b)1/2
- c)1/3
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
A man alternately tosses a coin and throws a dice beginning with the coin. The probability that he gets a head in the coin before he gets a 5 or 6 in the dice is
a)
3/4
b)
1/2
c)
1/3
d)
None of these
|
Ujwal Chawla answered |
Method to Solve :
Event and probability for coming head = H and 1/2
Event and probability for coming tail = T and 1/2
Event and probability for coming 5 or 6 = F and 1/3
Event and probability for not coming 5 or 6 = N and 2/3
Event sequences which are in our favour:
H
T N H
T N T N H
………. and so on
Sum all the sequences to get the probability:
1/2 + 1/2 x 2/3 x1/2 + 1/2 x 2/3 � 1/2 x 2/3 x 1/2 + …….
= 1/2 x [ 1 + 1/3 + (1/3)^2 + ….]
= 1/2 x 3/2
= 3/4
Out of 15 tickets marked 1, 2, 3, ... 15, three are drawn at random. The chance that numbers on them are in AP is- a)7/65
- b)8/455
- c)9/15
- d)9/39
Correct answer is option 'A'. Can you explain this answer?
Out of 15 tickets marked 1, 2, 3, ... 15, three are drawn at random. The chance that numbers on them are in AP is
a)
7/65
b)
8/455
c)
9/15
d)
9/39
|
Rithika Mishra answered |
Remember, if the total number of tickets is (2n + 1) and three (3) tickets are drawn at random; then the number of ways that numbers on the drawn tickets are in AP is n2.
Here 2n + 1 = 15
⇒ 2n = 14
⇒ n = 7
Here 2n + 1 = 15
⇒ 2n = 14
⇒ n = 7
A cubical dice has 3 on three faces, 2 on two faces and 1 on the 6 thth face .It is tossed twice. The chance that both the tosses show an even number is- a)1/4
- b)1/9
- c)1/36
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
A cubical dice has 3 on three faces, 2 on two faces and 1 on the 6 thth face .It is tossed twice. The chance that both the tosses show an even number is
a)
1/4
b)
1/9
c)
1/36
d)
none of these
|
|
Priya Mishra answered |
The correct option is B.
Since we have only even number 2,so when the dice is rolled twice we have (2,2) four times which means 2 on first dice pairs with both the twos on the second dice , similarly second two on first dice pairs with both the twos on the second dice
Hence the probability = 4/36 = 1/9
Three distinct numbers are selected from first 100 natural numbers. The probability that all the three numbers are divisible by 2 and 3 is- a)4/25
- b)4/35
- c)4/55
- d)4/1155
Correct answer is option 'D'. Can you explain this answer?
Three distinct numbers are selected from first 100 natural numbers. The probability that all the three numbers are divisible by 2 and 3 is
a)
4/25
b)
4/35
c)
4/55
d)
4/1155
|
Pragati Patel answered |
If a number is divisible by both 2 and 3, then it is a multiple of 6.
Now, I assume that if a number is chosen, it is not available for choosing again.
Now, multiples of 6 are 6, 12, 18,...., 96...i.e. 16 numbers
P(1st) => 16/100
P(2nd\1st is multiple of 6) => 15/99
P(3rd\1st and 2nd is multiple of 6) => 14/98
Hence, 16/100 x 15/99 x 14/98 => 4/25 x 5/33 x 1/7
i.e. 4/5 x 1/33 x 1/7 => 4/1155
Now, I assume that if a number is chosen, it is not available for choosing again.
Now, multiples of 6 are 6, 12, 18,...., 96...i.e. 16 numbers
P(1st) => 16/100
P(2nd\1st is multiple of 6) => 15/99
P(3rd\1st and 2nd is multiple of 6) => 14/98
Hence, 16/100 x 15/99 x 14/98 => 4/25 x 5/33 x 1/7
i.e. 4/5 x 1/33 x 1/7 => 4/1155
A person writes 4 letters and addresses 4 envelopes. If the letters are placed in the envelopes at random, then the probability that all letters are not placed in the right envelope is- a)1/24
- b)1
- c)0
- d)23/24
Correct answer is option 'D'. Can you explain this answer?
A person writes 4 letters and addresses 4 envelopes. If the letters are placed in the envelopes at random, then the probability that all letters are not placed in the right envelope is
a)
1/24
b)
1
c)
0
d)
23/24
|
Jaideep Sengupta answered |
n(s) = total number of ways of putting 4 letters in 4 envelopes.
= 4P4 = 4! = 24
No. of ways that all letters are placed in right envelope is 1.
∴ n(E)= No. of ways that all letters are not placed in right envelope.
(i.e. at least one letter is put in wrong envelope) = 24 - 1 = 23.
= 4P4 = 4! = 24
No. of ways that all letters are placed in right envelope is 1.
∴ n(E)= No. of ways that all letters are not placed in right envelope.
(i.e. at least one letter is put in wrong envelope) = 24 - 1 = 23.
and 
A student appears for test I, II and III. The student is successful if he passes either in test I and II or test I and III. The probability of the student passing in test I, II, III are p, q and 1/2 respectively. If the probability that the student is successful is 1/2, then - a)p = 1, q = 0
- b)p = 2/3, q = 1/2
- c)There are infinitely many values of p and q
- d)All of the above
Correct answer is option 'D'. Can you explain this answer?
A student appears for test I, II and III. The student is successful if he passes either in test I and II or test I and III. The probability of the student passing in test I, II, III are p, q and 1/2 respectively. If the probability that the student is successful is 1/2, then
a)
p = 1, q = 0
b)
p = 2/3, q = 1/2
c)
There are infinitely many values of p and q
d)
All of the above
|
Anisha Deshpande answered |
Let A, B and C be the events that the student is successful in test I, II and III respectively, then P(the student is successful)
P[A∩B∩C')∪(A∩B'∩C)∪(A∩B∩C)]
= P(A∩B∩C') + P(A∩B'∩C) + P(A∩B∩C)
P(A).P(B).P(C') + P(A)P(B')P(C) + P(A)P(B)P(C)
{∴ A, B, C are independent}
This equation has infinitely many values of p and q.
= P(A∩B∩C') + P(A∩B'∩C) + P(A∩B∩C)
P(A).P(B).P(C') + P(A)P(B')P(C) + P(A)P(B)P(C)
{∴ A, B, C are independent}
This equation has infinitely many values of p and q.
The sample space associated to thrown a dice is- a)(1, 2, 3)
- b)(1, 2, 3, 4, 5, 6)
- c)(1, 3, 5)
- d)(2, 4, 6)
Correct answer is option 'B'. Can you explain this answer?
The sample space associated to thrown a dice is
a)
(1, 2, 3)
b)
(1, 2, 3, 4, 5, 6)
c)
(1, 3, 5)
d)
(2, 4, 6)
|
Kamali Anandbabu answered |
The size of the sample space is the total number of possible outcomes. So one dice has 6 sides which has sample space of 1,2,3,4,5 and 6.
Both A and B throw a dice. The chance that B throws a number not less than that thrown by A is- a)1/2
- b)15/36
- c)21/36
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Both A and B throw a dice. The chance that B throws a number not less than that thrown by A is
a)
1/2
b)
15/36
c)
21/36
d)
none of these
|
|
Nandini Patel answered |
c) 21/36
If both A and B throw a dice, there are 36 possible outcomes in total: 6 possibilities for what A throws and 6 possibilities for what B throws. Of these 36 outcomes, 15 of them result in B throwing a number that is not less than the number thrown by A. Therefore, the probability that B throws a number not less than that thrown by A is 15/36, which is approximately 41.7%.
To find the probability of this event, we can consider each of the possible outcomes separately. For example, if A throws a 1 and B throws a 2, then B has thrown a number not less than that thrown by A. If A throws a 3 and B throws a 3, then B has thrown a number not less than that thrown by A. If A throws a 4 and B throws a 6, then B has thrown a number not less than that thrown by A. There are a total of 15 such outcomes.
The probability that B throws a number less than that thrown by A is 21/36, or approximately 58.3%. This is the complementary probability to the probability that B throws a number not less than that thrown by A. The complementary probability is the probability of the event not occurring. In this case, the event "B throws a number not less than that thrown by A" is not occurring if B throws a number less than that thrown by A. The sum of the probability of an event occurring and the probability of the event not occurring is always 1, so the probability of an event occurring plus the complementary probability of the event is always equal to 1.
A four digit number is formed by using the digits 1,2,3,4,5,6,7 without repetition. What is the probability that it’s divisible by 2?- a)3/7
- b)4/7
- c)2/7
- d)1/7
Correct answer is option 'A'. Can you explain this answer?
A four digit number is formed by using the digits 1,2,3,4,5,6,7 without repetition. What is the probability that it’s divisible by 2?
a)
3/7
b)
4/7
c)
2/7
d)
1/7
|
Anushka Ahuja answered |
The total number of possible four-digit numbers using the digits 1, 2, 3, 4, 5, 6, 7 without repetition is 7P4 = 7!/3! = 7*6*5*4 = 840.
To find the probability of the number being divisible by 4, we need to find the number of four-digit numbers using these digits that are divisible by 4.
A number is divisible by 4 if the last two digits form a number divisible by 4.
The possible two-digit numbers that are divisible by 4 are 12, 16, 24, 32, 36, 52, 56, 64, 72, and 76.
For each of these two-digit numbers, there are 5 choices for the first digit (excluding the last two digits used), and 4 choices for the remaining digit.
So, the total number of four-digit numbers using these digits that are divisible by 4 is 10 * 5 * 4 = 200.
Therefore, the probability of a random four-digit number using the digits 1, 2, 3, 4, 5, 6, 7 without repetition being divisible by 4 is 200/840 = 5/21.
To find the probability of the number being divisible by 4, we need to find the number of four-digit numbers using these digits that are divisible by 4.
A number is divisible by 4 if the last two digits form a number divisible by 4.
The possible two-digit numbers that are divisible by 4 are 12, 16, 24, 32, 36, 52, 56, 64, 72, and 76.
For each of these two-digit numbers, there are 5 choices for the first digit (excluding the last two digits used), and 4 choices for the remaining digit.
So, the total number of four-digit numbers using these digits that are divisible by 4 is 10 * 5 * 4 = 200.
Therefore, the probability of a random four-digit number using the digits 1, 2, 3, 4, 5, 6, 7 without repetition being divisible by 4 is 200/840 = 5/21.
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