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All questions of Probability for Grade 7 Exam
The probability of getting an even number, when a die is thrown once, is:- a)1/6
- b)1/2
- c)1/3
- d)5/6
Correct answer is option 'B'. Can you explain this answer?
The probability of getting an even number, when a die is thrown once, is:
a)
1/6
b)
1/2
c)
1/3
d)
5/6
|
Pooja Shah answered |
If we throw a die once, then possible outcomes (s), are
S = { 1, 2, 3, 4, 5, 6 }
⇒ n(E) = 6
(i) Let E be the favourable outcomes of getting an even number, then
E = { 2, 4, 6 }
⇒ n(S) = 3
S = { 1, 2, 3, 4, 5, 6 }
⇒ n(E) = 6
(i) Let E be the favourable outcomes of getting an even number, then
E = { 2, 4, 6 }
⇒ n(S) = 3
Two coins are tossed together. The probability of getting head on both the coins is- a)1/4
- b)3/4
- c)0
- d)1/2
Correct answer is option 'A'. Can you explain this answer?
Two coins are tossed together. The probability of getting head on both the coins is
a)
1/4
b)
3/4
c)
0
d)
1/2
|
Krishna Iyer answered |
The sample space for the event is : (H,H) (T,H) (H,T) (T,T)
Therefore total outcomes= 4
Therefore total outcomes= 4
Probability = 1/4
If a digit is chosen at random from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 then the probability that it is odd is
- a) 1/9
- b) 5/9
- c) 4/9
- d) 2/3
Correct answer is option 'B'. Can you explain this answer?
If a digit is chosen at random from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 then the probability that it is odd is
a)
1/9b)
5/9c)
4/9d)
2/3
|
Nisha Choudhury answered |
The even numbers out of these are= 2, 4, 6, 8
No. of even numbers = 4
Total numbers = 9
So probability of choosing an odd number will be
= No. of odd numbers/Total numbers
= 5 / 9
Similarly probability of choosing an even number will be
= No. of even numbers/Total numbers
= 4 / 9
If a letter of English alphabet is chosen at random, then the probability that the letter is a consonant is:- a)11/13
- b)5/26
- c)10/13
- d)21/26
Correct answer is option 'D'. Can you explain this answer?
If a letter of English alphabet is chosen at random, then the probability that the letter is a consonant is:
a)
11/13
b)
5/26
c)
10/13
d)
21/26
|
|
Naina Sharma answered |
The probability that the letter is a consonant = (26 - 5)/26 = 21/26.
Cards each marked with one of the numbers 4, 5, 6,...., 20 are placed in a box and mixed thoroughly. Onecard is drawn at random from the box. Then, the probability of getting an even prime number is- a)o
- b)1
- c)1/2
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Cards each marked with one of the numbers 4, 5, 6,...., 20 are placed in a box and mixed thoroughly. One
card is drawn at random from the box. Then, the probability of getting an even prime number is
a)
o
b)
1
c)
1/2
d)
None of these
|
|
Pooja Shah answered |
We have one and only even prime number which is 2. So no. of favourable outcomes= 0 and total no. of outcomes=17
Probability of getting 2 =0 this means that 2 cannot occur .
Probability of getting 2 =0 this means that 2 cannot occur .
The probability of an event that is certain to happen is- a)2
- b)1
- c)0
- d)-1
Correct answer is option 'B'. Can you explain this answer?
The probability of an event that is certain to happen is
a)
2
b)
1
c)
0
d)
-1
|
Ananya Das answered |
The probability of an event is a number describing the chance that the event will happen. An event that is certain to happen has a probability of 1. An event that cannot possibly happen has a probability of zero. If there is a chance that an event will happen, then its probability is between zero and 1.
Examples of Events:
- tossing a coin and it landing on heads
- tossing a coin and it landing on tails
- rolling a '3' on a die
- rolling a number > 4 on a die
- it rains two days in a row
- drawing a card from the suit of clubs
- guessing a certain number between 000 and 999 (lottery)
The probability of getting a prime number in single throw of a dice is:- a)Zero
- b)1/2
- c)1/4
- d)1/3
Correct answer is option 'B'. Can you explain this answer?
The probability of getting a prime number in single throw of a dice is:
a)
Zero
b)
1/2
c)
1/4
d)
1/3
|
|
Krishna Iyer answered |
Given, A dice is thrown once. So,
Total number of outcomes (n) = 6
Number of prime numbers = {2, 3,5}
So, Favorable number of outcomes (m) = 3
Thus, probability of getting a prime number = m/n = 3/6 = 1/2
If the probability of winning a game is 0.995, then the probability of losing is- a)0.05
- b)1
- c)0.005
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
If the probability of winning a game is 0.995, then the probability of losing is
a)
0.05
b)
1
c)
0.005
d)
None of the above
|
Sun Ray Institute answered |
Let P(winning the game)=0.995
Since the probability of two complimentary event sums to 1 so-
P(not winning the game) + P(winning the game )=1
=P(losing it )=1-P(winning the game )
=1-0.995=0.005
A card is drawn from a pack of 52 cards at random. The probability of getting either an ace or a king card is- a)4/13
- b)8/13
- c)3/13
- d)2/13
Correct answer is option 'D'. Can you explain this answer?
A card is drawn from a pack of 52 cards at random. The probability of getting either an ace or a king card is
a)
4/13
b)
8/13
c)
3/13
d)
2/13
|
|
Krishna Iyer answered |
Number of Total outcomes = 52
Number of aces and Number of kings = 4 + 4 = 8
Required Probability = 8/52 = 2/13
Number of aces and Number of kings = 4 + 4 = 8
Required Probability = 8/52 = 2/13
A die is thrown once, the probability of getting a prime number is- a)2/3
- b)1/3
- c)1/2
- d)1/6
Correct answer is option 'C'. Can you explain this answer?
A die is thrown once, the probability of getting a prime number is
a)
2/3
b)
1/3
c)
1/2
d)
1/6
|
|
Manish Singh answered |
The total no .of possible outcomes = 6
No .of favourable outcomes = 3 (2,3,5)
Let this event be E . p(E) = No . of favourable outcomes / total no .of outcomes
= 3/6 .
= 1/2
So , probability of getting a prime number is 1/2
No .of favourable outcomes = 3 (2,3,5)
Let this event be E . p(E) = No . of favourable outcomes / total no .of outcomes
= 3/6 .
= 1/2
So , probability of getting a prime number is 1/2
The probability of getting a number between 1 and 6 is
a) 1/6
b) 2/6
c) 3/4
d) 2/3
Correct answer is option 'D'. Can you explain this answer?
|
Gaurav Kumar answered |
Between 1 and 6, meaning not 1 or 6, then we have four favorable choices (2, 3, 4, 5) out of six possible outcomes (1, 2, 3, 4, 5, 6). Therefore your probability is 4/6 = 2/3.
Which of the following cannot be the probability of an event?- a)0.7
- b)15%
- c)2/3
- d)-1.5
Correct answer is option 'D'. Can you explain this answer?
Which of the following cannot be the probability of an event?
a)
0.7
b)
15%
c)
2/3
d)
-1.5
|
Rohan Kapoor answered |
D cannot be a probability as it has negative value
Which of the following cannot be the probability of an event?
- a)1/3
- b)0.1
- c)3%
- d)17/16
Correct answer is option 'D'. Can you explain this answer?
Which of the following cannot be the probability of an event?
a)
1/3
b)
0.1
c)
3%
d)
17/16
|
|
Rising Star answered |
Probability= possible outcomes/ total no.of outcomes
* however possible outcomes can never be greater than total no.of outcomes..that is why 17/16 is incorrect...
* Moreover the probability can't exceed from 1....
Read the following text and answer the following questions on the basis of the same:Rahul and Ravi planned to play Business (board game) in which they were supposed to use two dice.
Ravi got the first chance to roll the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 8?- a)1/26
- b)5/36
- c)1/18
- d)0
Correct answer is option 'B'. Can you explain this answer?
Read the following text and answer the following questions on the basis of the same:
Rahul and Ravi planned to play Business (board game) in which they were supposed to use two dice.
Ravi got the first chance to roll the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 8?
a)
1/26
b)
5/36
c)
1/18
d)
0
|
|
Naina Sharma answered |
The outcomes when two dice are thrown together are
= (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Total outcomes = 36
No. of outcomes when the sum is 8 = 5
Probability = 5/36
Which one of the following cannot be the probability of an event?- a)1.1
- b)0.1
- c)0.9
- d)5%
Correct answer is option 'A'. Can you explain this answer?
Which one of the following cannot be the probability of an event?
a)
1.1
b)
0.1
c)
0.9
d)
5%
|
|
Krishna Iyer answered |
If the probability is in percentage we divide it by 100
So d) option is 0.05.And we know If an event is impossible its probability is zero. Similarly, if an event is certain to occur, its probability is one. The probability of any event lies in between these values. It is called the range of probability and is denoted as 0 ≤ P (E) ≤ 1.And probability more than one means that favourable outcomes are more than total outcomes which is wrong.
So d) option is 0.05.And we know If an event is impossible its probability is zero. Similarly, if an event is certain to occur, its probability is one. The probability of any event lies in between these values. It is called the range of probability and is denoted as 0 ≤ P (E) ≤ 1.And probability more than one means that favourable outcomes are more than total outcomes which is wrong.
The probability that a non leap year selected at random will contain 53 Sunday's is- a)1/7
- b)2/7
- c)3/7
- d)5/7
Correct answer is option 'A'. Can you explain this answer?
The probability that a non leap year selected at random will contain 53 Sunday's is
a)
1/7
b)
2/7
c)
3/7
d)
5/7
|
|
Amit Kumar answered |
A non-leap year has 365 days
A year has 52 weeks. Hence there will be 52 Sundays for sure.
52 weeks = 52 x 7 = 364 days .
365– 364 = 1 day extra.
In a non-leap year there will be 52 Sundays and 1day will be left.
This 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday, friday, Saturday, Sunday.
Of these total 7 outcomes, the favourable outcomes are 1.
Hence the probability of getting 53 sundays = 1/7.
Cards each marked with one of the numbers 4, 5, 6 …20 are put in a box and mixed thoroughly. One card is drawn at random from the box. The probability of getting an even prime number is- a)0
- b)1/20
- c)1
- d)1/2
Correct answer is option 'A'. Can you explain this answer?
Cards each marked with one of the numbers 4, 5, 6 …20 are put in a box and mixed thoroughly. One card is drawn at random from the box. The probability of getting an even prime number is
a)
0
b)
1/20
c)
1
d)
1/2
|
|
Neha Patel answered |
Number of Favourable outcomes= 0
Total outcomes=17
Probability=0/17
Or probability=0
The probability that a leap year has 53 Sundays is- a)3/7
- b)2/7
- c)4/7
- d)1/7
Correct answer is option 'B'. Can you explain this answer?
The probability that a leap year has 53 Sundays is
a)
3/7
b)
2/7
c)
4/7
d)
1/7
|
|
Ananya Das answered |
There are 366 days in a leap year, i.e, 1 more than a normal year.
Now, 52 weeks make up 344 days (52 x 7 = 344)
That means that we already have 52 sundays for sure.
That means that we already have 52 sundays for sure.
Then, we are left with 2 days. Now, these days can be any from a pair of- mon-tues,tues-wed,wed-thurs,thurs-fri,fri-sat,sat-sun,sun-mon. Here favourable cases are sat-sun and sun-mon i.e, 2 cases and total number of cases is 7.
So, Probability=number of favourable cases/Total number of cases.
Therefore, Probability= 2/7.
If the probability of winning a game is 0.3, then the probability of losing it is:- a)0.3
- b)0.7
- c)0.2
- d)0.5
Correct answer is option 'B'. Can you explain this answer?
If the probability of winning a game is 0.3, then the probability of losing it is:
a)
0.3
b)
0.7
c)
0.2
d)
0.5
|
|
Amit Sharma answered |
Let the probability of losing the game be x
We know that sum of the the sum of probabilities is equal to 1
So
0.3+x=1
x=0.7
We know that sum of the the sum of probabilities is equal to 1
So
0.3+x=1
x=0.7
A card is drawn from a well-shuffled deck of 52 playing cards. The probability that the card will not be an ace card is
a)12/13b)1/4c)3/4d)1/13Correct answer is option 'A'. Can you explain this answer?
|
|
Ananya Das answered |
Total no. of outcomes=52
No. of ace cards=4
No. of non-ace cards=48
Probability of getting a non-ace card = No. of favourable outcomes / total no. of outcomes
=48/52=12/13
No. of ace cards=4
No. of non-ace cards=48
Probability of getting a non-ace card = No. of favourable outcomes / total no. of outcomes
=48/52=12/13
Two coins are tossed simultaneously once. Find the probability of getting at least one head and one tail.- a)1/4
- b)zero
- c)1/2
- d)1
Correct answer is option 'C'. Can you explain this answer?
Two coins are tossed simultaneously once. Find the probability of getting at least one head and one tail.
a)
1/4
b)
zero
c)
1/2
d)
1
|
|
Himaja Ammu answered |
Sample space ={H,T}{T,H}{H,H}{T,T} total no.of outcomes =4 no.of favourable outcomes =2 we know probability =no.of favourable outcomes/total no.of outcomes so probability =2/4=1/2
The probability that a randomly chosen number from one to twelve is a divisor of twelve is- a)1/6
- b)1/12
- c)1/2
- d)1/4
Correct answer is option 'D'. Can you explain this answer?
The probability that a randomly chosen number from one to twelve is a divisor of twelve is
a)
1/6
b)
1/12
c)
1/2
d)
1/4
|
|
Arun Yadav answered |
The answer is c.
No of favorable events i.e divisors from 1-12 = 1,2,3,4,6,12
No. of events = 12
Probabity that its a divisor of 12 = No. of favorable events/total no. of events
= 6/12 = 1/2
hence P(divisor of 12) = 1/2/ 0.5/0%
A die is thrown once. The probability of getting an even number and a multiple of 3 is- a)1/5
- b)1/3
- c)1/6
- d)1/2
Correct answer is option 'C'. Can you explain this answer?
A die is thrown once. The probability of getting an even number and a multiple of 3 is
a)
1/5
b)
1/3
c)
1/6
d)
1/2
|
Abhiram Malik answered |
Number of possible outcomes = {6} = 1
Number of Total outcomes = 6
∴ Required Probability = 1/6
Number of Total outcomes = 6
∴ Required Probability = 1/6
A bag has 9 red, 7 green and 4 blue balls. A student randomly selects a ball from the bag. The probability of not getting a blue ball is- a)4/5
- b)7/20
- c)1/5
- d)9/20
Correct answer is option 'A'. Can you explain this answer?
A bag has 9 red, 7 green and 4 blue balls. A student randomly selects a ball from the bag. The probability of not getting a blue ball is
a)
4/5
b)
7/20
c)
1/5
d)
9/20
|
Vikram Kapoor answered |
Total number of balls in the bag=9+4+7=20 balls
No. of favourable outcomes=not getting blue ball= getting either the red ball or green ball=9+7=16
Probability of not getting blue ball = no. of favourable outcomes/total number of outcomes=16/20=4/5
No. of favourable outcomes=not getting blue ball= getting either the red ball or green ball=9+7=16
Probability of not getting blue ball = no. of favourable outcomes/total number of outcomes=16/20=4/5
What is the probability of getting no head when two coins are tossed?a)1/4b)3/4c)1/2d)None of theseCorrect answer is 'A'. Can you explain this answer?
|
|
Amit Sharma answered |
Two coins are tossed simultaneously, so there are four possible outcomes ie, HH,HT,TT,TH
Total number of outcomes=4
Probability of getting no head=no. of favourable outcomes/ total no. of outcomes
=1/4
Total number of outcomes=4
Probability of getting no head=no. of favourable outcomes/ total no. of outcomes
=1/4
A letter of English alphabets is chosen at random. The probability that the letter chosen is a vowel is- a)1/26
- b)5/26
- c)4/26
- d)2/26
Correct answer is option 'B'. Can you explain this answer?
A letter of English alphabets is chosen at random. The probability that the letter chosen is a vowel is
a)
1/26
b)
5/26
c)
4/26
d)
2/26
|
Harshad Majumdar answered |
Number of consonants = 5
Number of total outcomes = 26
∴ Required Probability = 5/26
Number of total outcomes = 26
∴ Required Probability = 5/26
If an event cannot occur, then its probability is- a)1
- b)3/4
- c)1/2
- d)0
Correct answer is option 'D'. Can you explain this answer?
If an event cannot occur, then its probability is
a)
1
b)
3/4
c)
1/2
d)
0
|
|
Tanisha Singh answered |
If a event can not occur,then its favourable outcome will be 0 ,if u will divide 0 with any no. quotient will be zero.. therefore the probability of a event that can not occur will be 0
In a simultaneous throw of two coins the probability of getting at least one head is
a)1/2b)1/4c)3/4d)None of theseCorrect answer is option 'C'. Can you explain this answer?
|
|
Pooja Shah answered |
Two coins are simultaneously tossed.
So sample space={HH,HT,TH,TT}
No. of favourable outcomes=getting at least one head={HH,HT,TH}
=3
Total number of outcomes=4
Probability of getting at least one head=No. of favourable outcomes/Total number of outcomes
=3/4
So sample space={HH,HT,TH,TT}
No. of favourable outcomes=getting at least one head={HH,HT,TH}
=3
Total number of outcomes=4
Probability of getting at least one head=No. of favourable outcomes/Total number of outcomes
=3/4
The probability of getting a red and a king card is- a)5/26
- b)1/13
- c)7/26
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
The probability of getting a red and a king card is
a)
5/26
b)
1/13
c)
7/26
d)
None of these
|
|
Amit Kumar answered |
Let D be the favourable outcomes of getting red and a king, then n(D)=2 Therefore,
An urn contains lottery tickets numbered from 1 to 100. If a ticket is selected at random, then the probability that it is a perfect square is- a)0.1
- b)0.08
- c)0.09
- d)0.01
Correct answer is option 'A'. Can you explain this answer?
An urn contains lottery tickets numbered from 1 to 100. If a ticket is selected at random, then the probability that it is a perfect square is
a)
0.1
b)
0.08
c)
0.09
d)
0.01
|
Ujjwal Kumar answered |
Dude see....from 1 to 100...there is 10 perfect square number.
so, ur fav. number of out comes = 10.
nd number of total outcome = 100.
so....probability will be...no. of fav outcomes divided by number of total outcomes.
i.e., 10/100 => 1/10 => 0.1.
hope uh got it.
so, ur fav. number of out comes = 10.
nd number of total outcome = 100.
so....probability will be...no. of fav outcomes divided by number of total outcomes.
i.e., 10/100 => 1/10 => 0.1.
hope uh got it.
A bag contains cards which are numbered from 2 to 90. A card is drawn at random from the bag. The probability that it bears a two digit number is:- a)88/90
- b)88/92
- c)81/89
- d)89/90
Correct answer is option 'C'. Can you explain this answer?
A bag contains cards which are numbered from 2 to 90. A card is drawn at random from the bag. The probability that it bears a two digit number is:
a)
88/90
b)
88/92
c)
81/89
d)
89/90
|
|
Pooja Shah answered |
Total number of outcomes=90-1=89
No. of favourable outcomes = Total outcomes - One digit number =89-8=81
Probability of having a two digit number=Total number of outcomes/No. of favourable outcomes=81/89
No. of favourable outcomes = Total outcomes - One digit number =89-8=81
Probability of having a two digit number=Total number of outcomes/No. of favourable outcomes=81/89
Read the following text and answer the following questions on the basis of the same:On a weekend Rani was playing cards with her family. The deck has 52 cards.If her brother drew one card.
Find the probability of getting a red face card.- a)3/13
- b)1/13
- c)1/52
- d)1/4
Correct answer is option 'A'. Can you explain this answer?
Read the following text and answer the following questions on the basis of the same:
On a weekend Rani was playing cards with her family. The deck has 52 cards.If her brother drew one card.
Find the probability of getting a red face card.
a)
3/13
b)
1/13
c)
1/52
d)
1/4
|
|
Aditya Shah answered |
No. of face card = 13
Total no of cards = 52
Probability of getting a face card
= No. of face cards/Total no. of cards
= 12/52 = 3/13
The probability that a non-leap year selected at random will have 53 Mondays is- a)1/7
- b)7/52
- c)52/365
- d)45/52
Correct answer is option 'A'. Can you explain this answer?
The probability that a non-leap year selected at random will have 53 Mondays is
a)
1/7
b)
7/52
c)
52/365
d)
45/52
|
|
Anjana Khatri answered |
We know that there are 52 weeks in a year.
There are 7 days in a week
52 weeks will be to 7*52 = 364 days.
The remaining 1 day can be any day among Monday, Tuesday,..., Sunday.
Sample space has seven days as options.
So probability of getting 53 Sundays in a non-leap year is 1/7
Two coins are tossed simultaneously. The probability of getting atmost one head is- a)1/4
- b)1/2
- c)3/4
- d)1
Correct answer is option 'C'. Can you explain this answer?
Two coins are tossed simultaneously. The probability of getting atmost one head is
a)
1/4
b)
1/2
c)
3/4
d)
1
|
|
Pooja Shah answered |
Two coin tossed
Total sample space = {HH, HT, TH, TT}
Sample space for getting atmost one head = {TT, TH, HT}
P(s) = 3/4
Total sample space = {HH, HT, TH, TT}
Sample space for getting atmost one head = {TT, TH, HT}
P(s) = 3/4
Cards marked with numbers 1, 2, 3, ……….., 25 are placed in a box and mixed thoroughly and one card is drawn at random from the box. The probability that the number on the card is a multiple of 3 or 5 is
- a)8/25
- b)12/25
- c)4/25
- d)3/5
Correct answer is option 'B'. Can you explain this answer?
Cards marked with numbers 1, 2, 3, ……….., 25 are placed in a box and mixed thoroughly and one card is drawn at random from the box. The probability that the number on the card is a multiple of 3 or 5 is
a)
8/25
b)
12/25
c)
4/25
d)
3/5
|
Srishti Singh answered |
Harmeet tosses two coins simultaneously. The probability of getting at least one head is- a)1/2
- b)3/4
- c)2/3
- d)1/3
Correct answer is option 'B'. Can you explain this answer?
Harmeet tosses two coins simultaneously. The probability of getting at least one head is
a)
1/2
b)
3/4
c)
2/3
d)
1/3
|
Sneha Khroad2209 answered |
At least means minimum and that is 3/4
An event is very unlikely to happen. Its probability is closest to- a)0.0001
- b)0.001
- c)0.01
- d)0.1
Correct answer is option 'A'. Can you explain this answer?
An event is very unlikely to happen. Its probability is closest to
a)
0.0001
b)
0.001
c)
0.01
d)
0.1
|
|
Intelligent Nobita answered |
If a event is unlikely to happen means it's probability will be least .so,in the given options 0.001is clearly smallest . So option A is correct..
- a)1/2
- b)2/3
- c)1/3
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
a)
1/2
b)
2/3
c)
1/3
d)
None of these
|
|
Aryan Gupta. answered |
Baby Q 👆👆👆👆👆👆👆
nahi dekh rhaa... ... sorry bahenaa. wali baby.😎😎😎😎😎😉😉😉
A card is drawn at random from a pack of 52 playing cards. The probability of getting a face card is- a)1/4
- b)4/13
- c)3/13
- d)5/13
Correct answer is option 'C'. Can you explain this answer?
A card is drawn at random from a pack of 52 playing cards. The probability of getting a face card is
a)
1/4
b)
4/13
c)
3/13
d)
5/13
|
|
Anjana Khatri answered |
Clearly, there are 52 cards, out of which there are 12 face cards.
P (getting a face card) = 12/52=3/13.
A bag contains 4 red balls and 3 green balls. A ball is drawn at random. The probability of drawing a green ball is- a)1/7
- b)2/7
- c)3/7
- d)4/7
Correct answer is option 'D'. Can you explain this answer?
A bag contains 4 red balls and 3 green balls. A ball is drawn at random. The probability of drawing a green ball is
a)
1/7
b)
2/7
c)
3/7
d)
4/7
|
|
Gaurav Kumar answered |
Total number of outcomes=7
No. of favourable outcomes=3 (no. Of green balls)
Probability of getting a green ball=No. Of favourable outcomes/Total no. of outcomes=3/7
No. of favourable outcomes=3 (no. Of green balls)
Probability of getting a green ball=No. Of favourable outcomes/Total no. of outcomes=3/7
A bag contains 50 balls of which 2x are red, 3x are white and 5x are blue. A ball is selected at random. The probability that it is not white is
- a)2/3
- b)3/5
- c)7/10
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
A bag contains 50 balls of which 2x are red, 3x are white and 5x are blue. A ball is selected at random. The probability that it is not white is
a)
2/3
b)
3/5
c)
7/10
d)
none of these
|
|
Abhiram Malik answered |
Here
2x + 3x + 5x = 50 = ⇒ 10x = 50 ⇒ x = 5
Number of red balls = 2 x 5 = 10
Number of white balls = 3 x 5 = 15
Number of blue balls = 5 x 5 = 25
Now, Number of possible outcomes = 25 + 10 = 35
And Number of total outcomes = 50
∴ Required Probability = 35/50 = 7/10
2x + 3x + 5x = 50 = ⇒ 10x = 50 ⇒ x = 5
Number of red balls = 2 x 5 = 10
Number of white balls = 3 x 5 = 15
Number of blue balls = 5 x 5 = 25
Now, Number of possible outcomes = 25 + 10 = 35
And Number of total outcomes = 50
∴ Required Probability = 35/50 = 7/10
A piggy bank contains 100 fifty paise coins, 50 one rupee coins, 20 two rupee coins and 10 five rupee coins. One coin is drawn at random. The probability that the coin drawn will not be a five rupee coin is- a)8/9
- b)7/18
- c)5/9
- d)17/18
Correct answer is option 'D'. Can you explain this answer?
A piggy bank contains 100 fifty paise coins, 50 one rupee coins, 20 two rupee coins and 10 five rupee coins. One coin is drawn at random. The probability that the coin drawn will not be a five rupee coin is
a)
8/9
b)
7/18
c)
5/9
d)
17/18
|
|
Krishna Iyer answered |
Number of total coins = 100 + 50 + 20 + 10 = 180
Number of coins except five rupee coins = 180 – 10 = 170
∴ Required Probability = 170/180 = 17/18
Number of coins except five rupee coins = 180 – 10 = 170
∴ Required Probability = 170/180 = 17/18
In a single throw of a die, the probability of getting a multiple of 3 is
- A:
1/6
- B:
1/3
- C:
3/6
- D:
4/6
The answer is B.
1/6
1/3
3/6
4/6
|
Harshitha Das answered |
Given : A die is thrown once .
A die has 6 faces marked as 1, 2, 3, 4, 5 and 6.
If we throw one die then there possible outcomes are as follows: 1, 2, 3, 4, 5 and 6
Number of possible outcomes are = 6
Let E = Event of getting a getting a multiple of 3
Multiples of 3 are = 3, 6
Number of outcome favourable to E = 2
Probability (E) = Number of favourable outcomes / Total number of outcomes
P(E) = 2/6 = 1/3
Hence, the probability of getting a multiple of 3, P(E) = 1/3
The probability of getting one head is- a)1/2
- b)1/4
- c)3/4
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The probability of getting one head is
a)
1/2
b)
1/4
c)
3/4
d)
None of these
|
|
Pooja Shah answered |
Assume that is an equal chance of the coin landing on heads or tail (the coin is fair, not biased).
Probability of coin landing on head and then on tails=1/2 x 1/2=1/4
Probability of coin landing on tails and then on heads= 1/2 x 1/2 =1/4
Therefore, probability of getting one head and one tail in two coin tosses=1/4+1/4=1/2
From a well-shuffled pack of 52 cards, a card is drawn at random. The probability that it is a face card is:- a)2/13
- b)1/13
- c)4/13
- d)3/13
Correct answer is option 'D'. Can you explain this answer?
From a well-shuffled pack of 52 cards, a card is drawn at random. The probability that it is a face card is:
a)
2/13
b)
1/13
c)
4/13
d)
3/13
|
卩尺乇乇ㄒ卂爪 卩ㄚ卂尺乇 answered |
P(getting face card)
=(Total no.of face cards)/(Total no. of cards of shuffled deck i.e.52)
=12/52
=3/13.. (D)
A fair die is cast in the game of ‘Ludo’. The probability of getting a score greater than 6 is- a)1
- b)1/6
- c)zero
- d)2/3
Correct answer is option 'C'. Can you explain this answer?
A fair die is cast in the game of ‘Ludo’. The probability of getting a score greater than 6 is
a)
1
b)
1/6
c)
zero
d)
2/3
|
|
Ananya Das answered |
A fair dice has number 1,2,3,4,5,6 only . So there are no number greater than 6
No. of favourable outcomes=0
Total no. of outcomes=6
Probability of getting no. higher than 6=No. of favourable outcomes/Total no. of outcomes=0/6=0
No. of favourable outcomes=0
Total no. of outcomes=6
Probability of getting no. higher than 6=No. of favourable outcomes/Total no. of outcomes=0/6=0
Which of the following cannot be the probability of an event?- a)1.5
- b)3/5
- c)25%
- d)0.3
Correct answer is option 'A'. Can you explain this answer?
Which of the following cannot be the probability of an event?
a)
1.5
b)
3/5
c)
25%
d)
0.3
|
|
Rohit Sharma answered |
∵ Probability of any event cannot be more than 1.
∴ 1.5 can not be the probability of any event.
∴ (a) is the answer.
∴ 1.5 can not be the probability of any event.
∴ (a) is the answer.
Two fair coins are tossed simultaneously. Find the probability of Getting only one head- a)1/2
- b)1/3
- c)2/3
- d)1/4
Correct answer is option 'A'. Can you explain this answer?
Two fair coins are tossed simultaneously. Find the probability of Getting only one head
a)
1/2
b)
1/3
c)
2/3
d)
1/4
|
|
Arun Sharma answered |
When two coins are tossed, the sample space of possible outcomes would be the set
S = {HH, HT, TH, TT}
where H: the outcome is a “Head” and T: the outcome is a “Tail” and the probabilities associated with each of these outcomes are equal assuming the coins are fair.
When we say at most 1 “Head” outcome, it means that only 1 of the two coins show up a head or that the no “Head” outcome shows up.
If A is the event obtaining only 1 “Head” outcome, then the set of outcomes that makes up event A is
A = {(HT), (TH)}
Hence, the probability of event A is
P(A) = 2/4
A coin is tossed twice. The probability of getting both heads is- a)1/2
- b)1/3
- c)1/4
- d)1
Correct answer is option 'C'. Can you explain this answer?
A coin is tossed twice. The probability of getting both heads is
a)
1/2
b)
1/3
c)
1/4
d)
1
|
|
Pooja Shah answered |
Sample space - {HH, HT, TH, TT}
Number of total possible outcomes = 4
Number of favourable outcome (both heads) = 1
∴ Probability of getting both head = 1/4
Number of total possible outcomes = 4
Number of favourable outcome (both heads) = 1
∴ Probability of getting both head = 1/4
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