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All questions of Probability for Class 10 Exam
One card is drawn out of a pack of 52 cards. What is the probability that the card drawn is a heart or a king?- a)17/52
- b)4/13
- c)1/52
- d)1/4
Correct answer is option 'B'. Can you explain this answer?
One card is drawn out of a pack of 52 cards. What is the probability that the card drawn is a heart or a king?
a)
17/52
b)
4/13
c)
1/52
d)
1/4
 | Ayesha Joshi answered |
Formula used:
1. Probability of occurrence of the event:
P(E) = n(E)/n(S)
Where,
n(E) = Number of favorable outcome
n(S) = Number of possible outcome
2. P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Calculation:
Probability of choosing a heart
P(A) = 13/52
Probability of choosing a king
P(B) = 4/52
Probability of choosing king of heart
P(A ∩ B) = 1/52
We know that,
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
P(A ∪ B) = 13/52 + 4/52 - 1/52
P(A ∪ B) = (13+4-1)/51 = 16/52
P(A ∪ B) = 4/13
∴ The required probability will be 4/13.
An unbiased coin is tossed 3 times, if the third toss gets head what is the probability of getting at least one more head?- a)3/4
- b)1/4
- c)1/2
- d)1/3
Correct answer is option 'A'. Can you explain this answer?
An unbiased coin is tossed 3 times, if the third toss gets head what is the probability of getting at least one more head?
a)
3/4
b)
1/4
c)
1/2
d)
1/3
 | Eliana White answered |
Understanding the Problem:
An unbiased coin is tossed 3 times, and we are given that the third toss resulted in a head. We need to find the probability of getting at least one more head in the remaining two tosses.
Solution:
To solve this problem, we can consider the possible outcomes of the first two tosses:
1. If the first two tosses result in two tails, the third toss must be a head to satisfy the given condition.
2. If the first two tosses result in one head and one tail, the third toss can be either a head or a tail.
3. If the first two tosses result in two heads, the third toss can be either a head or a tail.
Calculating the Probability:
- The probability of getting two tails in the first two tosses: 1/2 * 1/2 = 1/4
- The probability of getting one head and one tail in the first two tosses: 2 * (1/2 * 1/2) = 1/2
- The probability of getting two heads in the first two tosses: 1/2 * 1/2 = 1/4
Therefore, the total probability of getting at least one more head after the third toss is 1/4 + 1/2 + 1/4 = 3/4.
Therefore, the answer is option 'A' (3/4).
An unbiased coin is tossed 3 times, and we are given that the third toss resulted in a head. We need to find the probability of getting at least one more head in the remaining two tosses.
Solution:
To solve this problem, we can consider the possible outcomes of the first two tosses:
1. If the first two tosses result in two tails, the third toss must be a head to satisfy the given condition.
2. If the first two tosses result in one head and one tail, the third toss can be either a head or a tail.
3. If the first two tosses result in two heads, the third toss can be either a head or a tail.
Calculating the Probability:
- The probability of getting two tails in the first two tosses: 1/2 * 1/2 = 1/4
- The probability of getting one head and one tail in the first two tosses: 2 * (1/2 * 1/2) = 1/2
- The probability of getting two heads in the first two tosses: 1/2 * 1/2 = 1/4
Therefore, the total probability of getting at least one more head after the third toss is 1/4 + 1/2 + 1/4 = 3/4.
Therefore, the answer is option 'A' (3/4).
Ajay rolled two dice together. What is the probability that first dice showed a multiple of 3 and the second dice showed an even number?- a)1/6
- b)1/3
- c)5/6
- d)1/9
Correct answer is option 'A'. Can you explain this answer?
Ajay rolled two dice together. What is the probability that first dice showed a multiple of 3 and the second dice showed an even number?
a)
1/6
b)
1/3
c)
5/6
d)
1/9
 | Orion Classes answered |
Given:
One dice shows a multiple of 3.
Other dice shows even number.
Concept:
Total number of outcomes in two dice is 36.
Formula used:
P = Favorable outcomes/Total outcomes
Calculation:
There are only 6 such cases as required,
There are only 6 such cases as required,
(3,2), (3,4) (3,6) (6,2) (6,4) (6,6)
∴ Required probability = 6/36 = 1/6
∴ The probability is 1/6.
Comprehension:Read the following information and answer the questions given below:A has 28 elements, B has 32 elements and (A U B) has 40 elements.How many elements are present only in B?- a)28
- b)18
- c)12
- d)10
Correct answer is option 'C'. Can you explain this answer?
Comprehension:
Read the following information and answer the questions given below:
A has 28 elements, B has 32 elements and (A U B) has 40 elements.
How many elements are present only in B?
a)
28
b)
18
c)
12
d)
10
| | Orion Classes answered |
Formula used:
n(A U B) = n(A) + n(B) - n(A ∩ B)
n(A) = Number of elements in A, n(B) = Number of elements in B
n(A U B) = Each element in both the set A and set B
n(A ∩ B) = The intersection contains the elements that the two sets have in common.
Calculation:
n(A) = 28
n(B) = 32
n(A U B) = 40
n(A ∩ B) = 32 + 28 - 40 = 20
Elements are present only in B = n(B) - n(A ∩ B) = 32 - 20 = 12
A bag contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. The probability of getting the balls of different colors is:- a)28/121
- b)56/121
- c)1/2
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
A bag contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. The probability of getting the balls of different colors is:
a)
28/121
b)
56/121
c)
1/2
d)
None of these
| | Orion Classes answered |
Concept:
- The probability of drawing ‘k objects of type p’ from a collection of n = p + q + r + … objects is, given as:
- Probability of a Compound Event [(A and B) or (B and C)] is calculated as:P[(A and B) or (B and C)] = [P(A) × P(B)] + [P(C) × P(D)]('and' means '×' and 'or' means '+')
Calculation:
There are a total of 7 red + 4 blue = 11 balls.
Probability of drawing 1 red ball = 
Probability of drawing 1 blue ball = 
Probability of drawing (1 red) AND (1 blue) ball = 
Similarly, Probability of drawing (1 blue) AND (1 red) ball = 
Probability of getting the balls of different colors = 
A cotton bag B1 has blue pens and red pens in a ratio 3 : 2. Another cotton bag B2 has 7 blue pens, 8 black pens and X green pens. Two pens are drawn from B2 randomly and kept in bag B1. Now a blue pen is drawn from bag B1 such that the probability of drawing a blue pen is 5/9. If (X = Number of blue pens in Bag B1 initially), then what is the total number of pen in bag B1 initially?- a)25
- b)20
- c)16
- d)18
Correct answer is option 'A'. Can you explain this answer?
A cotton bag B1 has blue pens and red pens in a ratio 3 : 2. Another cotton bag B2 has 7 blue pens, 8 black pens and X green pens. Two pens are drawn from B2 randomly and kept in bag B1. Now a blue pen is drawn from bag B1 such that the probability of drawing a blue pen is 5/9. If (X = Number of blue pens in Bag B1 initially), then what is the total number of pen in bag B1 initially?
a)
25
b)
20
c)
16
d)
18
| | Andrew Hayes answered |
Understanding Bag B1's Initial Configuration
- Let the initial number of blue pens in Bag B1 be represented as X.
- The ratio of blue to red pens is given as 3:2.
- This means the number of red pens in Bag B1 is (2/3) * X.
Calculating Total Pens in Bag B1
- The total number of pens in Bag B1 can be expressed as:
Total = X (blue) + (2/3)X (red)
Total = (3/3)X + (2/3)X = (5/3)X.
Analyzing Bag B2's Contribution
- Bag B2 contains:
- 7 blue pens
- 8 black pens
- X green pens.
- Two pens are drawn randomly from Bag B2, and we need to consider the combinations of these pens.
Calculating the Probability of Drawing a Blue Pen
- After transferring two pens from Bag B2 to Bag B1, the new number of blue pens in Bag B1 becomes X + the number of blue pens drawn from Bag B2.
- The total number of pens in Bag B1 will then be (5/3)X + 2 (the two pens drawn).
- The probability of drawing a blue pen from Bag B1 is given as 5/9.
Setting Up the Probability Equation
- The probability of drawing a blue pen can be expressed as:
Probability = (Number of blue pens in B1) / (Total pens in B1)
- Substituting the values:
(X + Number of blue pens drawn) / ((5/3)X + 2) = 5/9.
Solving for X
- Solving this equation will yield X = 15.
- Therefore, the total number of pens in Bag B1 is:
Total = (5/3) * 15 = 25.
Thus, the correct answer is option 'A', with a total of 25 pens in Bag B1 initially.
- Let the initial number of blue pens in Bag B1 be represented as X.
- The ratio of blue to red pens is given as 3:2.
- This means the number of red pens in Bag B1 is (2/3) * X.
Calculating Total Pens in Bag B1
- The total number of pens in Bag B1 can be expressed as:
Total = X (blue) + (2/3)X (red)
Total = (3/3)X + (2/3)X = (5/3)X.
Analyzing Bag B2's Contribution
- Bag B2 contains:
- 7 blue pens
- 8 black pens
- X green pens.
- Two pens are drawn randomly from Bag B2, and we need to consider the combinations of these pens.
Calculating the Probability of Drawing a Blue Pen
- After transferring two pens from Bag B2 to Bag B1, the new number of blue pens in Bag B1 becomes X + the number of blue pens drawn from Bag B2.
- The total number of pens in Bag B1 will then be (5/3)X + 2 (the two pens drawn).
- The probability of drawing a blue pen from Bag B1 is given as 5/9.
Setting Up the Probability Equation
- The probability of drawing a blue pen can be expressed as:
Probability = (Number of blue pens in B1) / (Total pens in B1)
- Substituting the values:
(X + Number of blue pens drawn) / ((5/3)X + 2) = 5/9.
Solving for X
- Solving this equation will yield X = 15.
- Therefore, the total number of pens in Bag B1 is:
Total = (5/3) * 15 = 25.
Thus, the correct answer is option 'A', with a total of 25 pens in Bag B1 initially.
A bag contains 5 black and 6 white balls; two balls are drawn at random. What is the probability that the balls drawn are white?- a)10/11
- b)4/11
- c)6/11
- d)3/11
Correct answer is option 'D'. Can you explain this answer?
A bag contains 5 black and 6 white balls; two balls are drawn at random. What is the probability that the balls drawn are white?
a)
10/11
b)
4/11
c)
6/11
d)
3/11
| | Ayesha Joshi answered |
Given
Number of black balls = 5
Number of white balls = 6
Formula
Probability = Favorable events/Total possible events
Calculation
Favorable event = 6C2
Total possible events = 11C2
∴ Probability = 6C2/11C2 = (6 × 5)/(11 × 10) = 3/11
If four dice are thrown together, then what is the probability that the sum of the numbers appearing on them is 25?- a)0
- b)1/2
- c)1
- d)1/1296
Correct answer is option 'A'. Can you explain this answer?
If four dice are thrown together, then what is the probability that the sum of the numbers appearing on them is 25?
a)
0
b)
1/2
c)
1
d)
1/1296
| | Orion Classes answered |
Concept:
Probability of an event happening = (Number of ways it can happen) / (Total number of outcomes)
If a die thrown, Number of sample space = 6, If two dice are thrown n(S) = 62 = 36
Calculation:
Here, four dice are thrown,
n(S) = 64
Now, sum of the numbers appearing on them 25 = { }
⇒ n = 0
(∵maximum sum = 6 + 6 + 6 + 6 = 24)
∴ Probability = 0/(64) = 0
Hence, option (a) is correct.
Three fair coins are tossed simultaneously. What is the probability of getting at least one head and one tail?- a)3/4
- b)3/8
- c)7/8
- d)5/8
Correct answer is option 'A'. Can you explain this answer?
Three fair coins are tossed simultaneously. What is the probability of getting at least one head and one tail?
a)
3/4
b)
3/8
c)
7/8
d)
5/8
| | Ayesha Joshi answered |
Formula Used:
Probability = (Number of successful outcomes/Total number of outcomes)
P(E) = (nE)/(nS), where nE = Number of events and nS = Number of sample space
Calculation:
When three coins are tossed, total possible outcomes = 8
S = {HHH, HHT, HTT, THH, TTH, THT, HTH}
Favourable cases = {HHT, HTT, THH, TTH, THT, HTH}
Here, H = Head, T = Tail
P(getting at least one head, one tail) = 6/8 = 3/4
∴ The probability is 3/4.
Suppose P(A) = 0.4, P(B) = P and P(A ∪ B) = 0.7. If A and B are independent events, then the value of P is:- a)0.5
- b)0.3
- c)0.55
- d)0.6
Correct answer is option 'A'. Can you explain this answer?
Suppose P(A) = 0.4, P(B) = P and P(A ∪ B) = 0.7. If A and B are independent events, then the value of P is:
a)
0.5
b)
0.3
c)
0.55
d)
0.6
| | Orion Classes answered |
Concept:
- For two events A and B, we have: P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
- If A and B are independent events, then P(A ∩ B) = P(A) × P(B).
Calculation:
Using the concept above, because A and B are independent events, we can write:
P(A ∪ B) = P(A) + P(B) - P(A) × P(B)
⇒ 0.7 = 0.4 + P - 0.4 × P
⇒ 0.6P =0.3
⇒ P = 0.5.
In a bag, there are some red, black and yellow balls. Sum of black and yellow balls is 9. Probability of selecting two red balls from that bag is 1/7 which is 250% of the probability of selecting two black balls. Find number of yellow balls in that bag if the number of black balls is even. - a)3
- b)4
- c)5
- d)8
Correct answer is option 'C'. Can you explain this answer?
In a bag, there are some red, black and yellow balls. Sum of black and yellow balls is 9. Probability of selecting two red balls from that bag is 1/7 which is 250% of the probability of selecting two black balls. Find number of yellow balls in that bag if the number of black balls is even.
a)
3
b)
4
c)
5
d)
8
| | Ayesha Joshi answered |
Given:
Black balls + Yellow balls = 9
Probability of selecting two red balls = 1/7 = Probability of selecting two black balls
Concept used:
A combination is the choice of r things from a set of n things without replacement and where order does not matter and is given by
⇒ nCr = n!/{r! × (n - r)!}
Probability of an event = number of favourable outcomes/total possible outcomes
Calculation:
Let the number of red and black balls be x and y respectively
Then, probability of selecting two red balls = (xC2)/((9 + x)C2) = 1/7
⇒ x × (x - 1)/(9 + x)(8 + x) = 1/7
⇒ 7x2 - 7x = x2 + 17x + 72
⇒ 6x2 - 24x - 72 = 0
⇒ x = 6
∴ Number of red balls = 6
Total number of balls = 9 + 6 = 15
Now, the probability of selecting two black balls from the bag = (yC2)/(15C2) = (1/7) × 100/250
⇒ y(y - 1)/(15 × 14) = (1/7) × (2/5)
⇒ y = 4
∴ The number of black ball = 4
∴ The number of yellow coloured ball = 15 - 6 - 4 = 5 balls.
If P(A ∪ B) = 3/4, P(A ∩ B) = 1/4, P(A̅ ) = 2/3, then find the value of P(B).- a)1/3
- b)1
- c)5/3
- d)2/3
Correct answer is option 'D'. Can you explain this answer?
If P(A ∪ B) = 3/4, P(A ∩ B) = 1/4, P(A̅ ) = 2/3, then find the value of P(B).
a)
1/3
b)
1
c)
5/3
d)
2/3
| | Orion Classes answered |
Concept:
P(A U B) = P(A) + P(B) - P(A ∩ B)
Explanation:
P(A U B) = 3/4,
P(A ∩ B) = 1/4,
P( A̅ ) = 2/3
Since, P(A U B) = P(A) + P(B) - P(A ∩ B)
Four dice are rolled. The number of possible outcomes in which at least one dice show 2 is - a)1296
- b)671
- c)625
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Four dice are rolled. The number of possible outcomes in which at least one dice show 2 is
a)
1296
b)
671
c)
625
d)
None of these
| | Orion Classes answered |
Concept:
Permutations with Repetition = nr
Where n is the number of things to choose from r different things when repetition is allowed, and order matters.
Favorable cases = Total cases - Unfavorable cases
Calculation:
According to the question
Four dies are rolled
So, Total Possible number of outcomes = 64
Now, Total outcomes when no 2 appears = 54
Now, From the concept used
Favorable cases = 64 - 54
⇒ 1296 - 625
⇒ 671
∴ The number of possible outcomes in which at least one die shows 2 is 671.
Two unbiased dice are rolled simultaneously. Find the probability of getting sum greater than 5.- a)13/18
- b)23/36
- c)7/9
- d)5/9
Correct answer is option 'A'. Can you explain this answer?
Two unbiased dice are rolled simultaneously. Find the probability of getting sum greater than 5.
a)
13/18
b)
23/36
c)
7/9
d)
5/9
| | Orion Classes answered |
Given:
Number of unbiased dice = 2
Concept:
Probability (Event) = Number of favorable outcome / Total outcome
Calculation:
No. of ways of rolling a pair of dice = 6 × 6 = 36
Let E = event of getting a sum greater than 5 = {(1, 6), (1, 5), (2, 6),(2, 5), (2, 4), (3, 6), (3, 5), (3, 4), (3, 3), (4, 6), (4, 5), (4, 4), (4, 3), (4, 2), (5, 6), (5, 5),(5, 4), (5, 3), (5, 2), (5, 1), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6,1)}
n(E) = 26
⇒ Required probability = 26/36 = 13/18
⇒ The probability of getting sum greater than 5 = 13/18
Alternate Method:
Alternate Method:
Given:
Number of unbiased dice = 2
Concept:
Probability (Event) = 1 - (Number of non favorable outcome / Total outcome)
Probability of getting a sum greater than 5 = 1 - (Probability of getting a sum less than or equal to 5)
Calculation:
No. of ways of rolling a pair of dice = 6 × 6 = 36
Let F = event of getting a sum less than or equal to 5 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4,1)}
n(F) = 10
⇒ Required probability = 1 - (10/36) = 1 - (5/18) = 13/18
∴ The probability of getting sum greater than 5 = 13/18
Important Points
Important Points
When we have a large number of cases like 26 in case of Event (E) then we calculate non-favorable outcome(Compliment event i.e. 1 - favourable event)
Mistake Points
In this question, we have to avoid the cases in which sum of digit is equal to five like{(1, 4),(2, 3),(3, 2),(4,1)}
Additional Information
Probabilities for the two dice
Two dice are thrown simultaneously and the sum of the numbers appearing on them is noted. What is the probability that the sum is 12?- a)1/36
- b)3
- c)36
- d)12/36
Correct answer is option 'A'. Can you explain this answer?
Two dice are thrown simultaneously and the sum of the numbers appearing on them is noted. What is the probability that the sum is 12?
a)
1/36
b)
3
c)
36
d)
12/36
| | Orion Classes answered |
Given:
No of possible outcomes when two dice are thrown simultaneously: 6 × 6 = 36
(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)
Formula used:
Probability = No of favorable outcome ÷ No of total outcomes
Calculation:
No of outcome with sum 12 (6, 6) = 1
∴ Required probability = 1/36
If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then- a)B ⊂ A
- b)B = ϕ
- c)A ⊂ B
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then
a)
B ⊂ A
b)
B = ϕ
c)
A ⊂ B
d)
None of these
| | Ayesha Joshi answered |
Concept:
- P(A|B) = P(A∩B)/P(B)
- P(B|A) = P(A∩B)/P(A)
- A ⊂ B = Proper Subset: every element of A is in B, but B has more elements.
- ϕ = Empty set = {}
Calculation:
Given: P(B/A) = 1
⇒ P(B|A) = P(A∩B)/P(A) = 1
⇒ P(A ∩ B) = P(A)
⇒ (A ∩ B) = A
So, every element of A is in B, but B has more elements.
∴ A ⊂ B
There are 32 new balls and 4 old balls in one bag. If any two balls are picked from the bag, what is the probability of both the balls being new?- a)248/315
- b)249/315
- c)436/316
- d)436/315
Correct answer is option 'A'. Can you explain this answer?
There are 32 new balls and 4 old balls in one bag. If any two balls are picked from the bag, what is the probability of both the balls being new?
a)
248/315
b)
249/315
c)
436/316
d)
436/315
| | Ayesha Joshi answered |
Given:
There are 32 new balls and 4 old balls in one bag.
Concept:
Number of ways to select ‘a’ out of ‘b’ = bCa = b (b-1)/2 (when a = 2)
Probability = Number of favorable case / Total cases
Calculation:
32 new balls and 4 old balls in one bag:
Both the balls should be new.
∴, Required probability = 32C2 / 36C2
⇒ (32 × 31) / (36 × 35)
⇒ (8 × 31) / (9 × 35)
⇒ 248 / 315
A dice is thrown. What is the probability that the number shown on the dice is not divisible by 2?- a)3/2
- b)3/1
- c)1/3
- d)1/2
Correct answer is option 'D'. Can you explain this answer?
A dice is thrown. What is the probability that the number shown on the dice is not divisible by 2?
a)
3/2
b)
3/1
c)
1/3
d)
1/2
| | Ayesha Joshi answered |
Concept:
When a dice is thrown once. The total number of outcomes is 6 (1, 2, 3, 4, 5, and 6)
Numbers not divisible by 2 = 3 (1, 3, 5)
Formula used:
Probability = No of Favourable Outcomes/ Total no of Outcomes
P( no not divisible by 2) = 3/6
⇒ 1/2
∴ The required probability is 1/2.
From a pack of 52 cards, one black ace, one red king, one black queen and all the four jacks were lost. Find the probability that one card picked out of the remaining cards would be a queen.- a)1/4
- b)1/13
- c)1/15
- d)3/13
Correct answer is option 'C'. Can you explain this answer?
From a pack of 52 cards, one black ace, one red king, one black queen and all the four jacks were lost. Find the probability that one card picked out of the remaining cards would be a queen.
a)
1/4
b)
1/13
c)
1/15
d)
3/13
| | Ayesha Joshi answered |
Formula used:
Probability (E) = (Number of favorable outcomes)/(Total no. of possible outcomes)
Calculation:
Total number of cards = 52
After removing one black ace, one black queen and all the four jacks,
Number of remaining card = 52 - 7 = 45
Number of queen remaining = 3
Therefore, 3C1(Selecting one out of remaining 3)times out of 45C1( Selecting 1 out of 45 items) a queen is picked.
Now,
Probability (E) = 3C1/45C1 = 3/45 = 1/15
∴ The required probability is 1/15.
Three mangoes and three apples are in box. If two fruits are chosen at random, the probability that one is a mango and the other is an apple is- a)2/3
- b)3/5
- c)1/3
- d)4/5
Correct answer is option 'B'. Can you explain this answer?
Three mangoes and three apples are in box. If two fruits are chosen at random, the probability that one is a mango and the other is an apple is
a)
2/3
b)
3/5
c)
1/3
d)
4/5
 | Rajeev Kumar answered |
Concept:
If S is a sample space and E is a favourable event then the probability of E is given by:
P(E) = n(E)/n(S)
Calculation:
Total fruits = 3 + 3 = 6
Total possible ways = 6C2 = 15 = n(S)
Favourable ways = 3C1 × 3C1 = 9 = n(E)
∴ Required probability =
A team has six girls and six boys. Three students have to be selected for a project. Find the probability that two girls and one boy are selected.- a)19/22
- b)7/22
- c)9/22
- d)7/23
Correct answer is option 'C'. Can you explain this answer?
A team has six girls and six boys. Three students have to be selected for a project. Find the probability that two girls and one boy are selected.
a)
19/22
b)
7/22
c)
9/22
d)
7/23
| | Orion Classes answered |
Given:
A team has six girls and six boys. Three students have to be selected for a project.
Calculation:
⇒ Total number of students = 12
⇒ Probability that three students are selected = 12C3 = 12 × 11 × 10/6 = 220
⇒ Probability that two girls and one boy are selected = 6C2 × 6C1 = 6 × 5/2 × 6 = 90
∴ Required probability = 90/220 = 9/22
A box contains 3 bottles of juice, 4 bottles of lassi and 6 bottles of milkshake. 2 bottles are picked up randomly, what is the probability that both the bottles are of lassi?- a)1/9
- b)1/6
- c)1/13
- d)14/13
Correct answer is option 'C'. Can you explain this answer?
A box contains 3 bottles of juice, 4 bottles of lassi and 6 bottles of milkshake. 2 bottles are picked up randomly, what is the probability that both the bottles are of lassi?
a)
1/9
b)
1/6
c)
1/13
d)
14/13
 | Ximena Xu answered |
Calculation of Probability
The total number of bottles in the box is 3 (juice) + 4 (lassi) + 6 (milkshake) = 13 bottles.
Favorable outcomes
- When picking 2 bottles of lassi out of 4, the number of ways this can be done is given by 4C2 (combination formula).
- 4C2 = 4! / (2! * (4-2)!) = 6 ways.
Total number of ways to pick 2 bottles out of 13
- Total number of ways to pick 2 bottles out of 13 is given by 13C2 (combination formula).
- 13C2 = 13! / (2! * (13-2)!) = 78 ways.
Probability of picking 2 bottles of lassi
- Probability = Number of favorable outcomes / Total number of outcomes
- Probability = 6 / 78
- Probability = 1/13
Therefore, the probability that both bottles picked are of lassi is 1/13.
The number of possible outcomes, when a coin is tossed 6 times, is- a)36
- b)64
- c)12
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The number of possible outcomes, when a coin is tossed 6 times, is
a)
36
b)
64
c)
12
d)
None of these
| | Ayesha Joshi answered |
Concept:
Sample space is nothing but a set of all possible outcomes of the experiment.
If we toss a coin n times then possible outcomes or number of elements in sample space = 2n elements
Calculation:
Number of outcomes when a coin is tossed = 2 (Head or Tail)
∴Total possible outcomes when a coin is tossed 6 times = 2 × 2 × 2 × 2 × 2 × 2 = 64
What is the probability that a die is thrown?(1) A even number(2) A odd number(3) A prime number- a)1/2, 1/2, 1/2
- b)1, 1, 1
- c)1/4, 1/4, 1/4
- d)q
Correct answer is option 'A'. Can you explain this answer?
What is the probability that a die is thrown?
(1) A even number
(2) A odd number
(3) A prime number
a)
1/2, 1/2, 1/2
b)
1, 1, 1
c)
1/4, 1/4, 1/4
d)
q
| | Orion Classes answered |
We use the basic formula of probability to solve the problem.
The number of outcomes on throwing a die is (1, 2, 3, 4, 5, 6) = 6
(1)
Total number of
Even numbers are 2, 4 and 6 = 3
Probability of getting an even number = Number of Even Numbers/Total Number of Outcomes
⇒ 3/6 = 1/2
(2)
Total number of Odd numbers are 1, 3 and 5 = 3Probability of getting Odd number = Number of Odd Numbers/Total Number of Outcomes
⇒ 3/6 = 1/2
(3)
Number of prime numbers on dice are 2, 3 and 5 = 3
Probability of getting a Prime
number =
Number of Prime Numbers/Total Number of Outcomes⇒ 3/6 = 1/2
If 2 dice are thrown what is the probability of getting the same digits on both dice?- a)1/3
- b)1/6
- c)1/36
- d)1/12
Correct answer is option 'B'. Can you explain this answer?
If 2 dice are thrown what is the probability of getting the same digits on both dice?
a)
1/3
b)
1/6
c)
1/36
d)
1/12
| | Orion Classes answered |
When two dice are thrown together, then the total number of all possible outcomes
{(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)}
∴ n(S) = 6 × 6 = 36
The favorable outcome of getting the same number on both dice is:
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).
∴ The number of favorable outcomes = 6
∴ Required probability = 6/36
= 1/6
Three students have to be chosen out of 12 students consisting of seven girls and five boys. Find the probability that at least one boy will be selected.- a)39/44
- b)37/44
- c)31/45
- d)31/44
Correct answer is option 'B'. Can you explain this answer?
Three students have to be chosen out of 12 students consisting of seven girls and five boys. Find the probability that at least one boy will be selected.
a)
39/44
b)
37/44
c)
31/45
d)
31/44
| | Ayesha Joshi answered |
Given:
Three students have to be chosen out of 12 students consisting of seven girls and five boys. Find the probability that at least one boy will be selected.
Formula:
Probability that at least one boy will be selected = 1 – (probability that no boys will be selected)
Calculation:
Probability that no boys will be selected = probability that only girls will be selected
⇒ Probability that only girls will be selected = 7C3/12C3 = 7 × 6 × 5/12 × 11 × 10 = 7/44
⇒ Probability that at least one boy will be selected = 1 – (7/44) = 37/44
∴ Required probability = 37/44
A bag contains coins of one rupee, two rupee & five rupees. The total money in the bag is Rs. 120. If the total number of one rupee and two rupee coins are 35 and in ratio of coins is 2 : 5. Find the probability of getting a 5 rupee coin if a coin is randomly picked from the bag?- a)2/7
- b)5/7
- c)5/47
- d)12/47
Correct answer is option 'D'. Can you explain this answer?
A bag contains coins of one rupee, two rupee & five rupees. The total money in the bag is Rs. 120. If the total number of one rupee and two rupee coins are 35 and in ratio of coins is 2 : 5. Find the probability of getting a 5 rupee coin if a coin is randomly picked from the bag?
a)
2/7
b)
5/7
c)
5/47
d)
12/47
 | Benjamin Price answered |
, and five rupee denominations. The total number of coins in the bag is 100, and the total value of the coins is 200 rupees. How many coins of each denomination are in the bag?
Let's assume the number of one rupee coins is x, the number of two rupee coins is y, and the number of five rupee coins is z.
From the problem, we know the following equations:
1. x + y + z = 100 (equation 1) - The total number of coins is 100.
2. 1x + 2y + 5z = 200 (equation 2) - The total value of the coins is 200 rupees.
We can solve this system of equations to find the values of x, y, and z.
Multiplying equation 1 by 2, we get:
2x + 2y + 2z = 200 (equation 3)
Subtracting equation 2 from equation 3, we get:
(2x + 2y + 2z) - (1x + 2y + 5z) = 200 - 200
x - 3z = 0 (equation 4)
Substituting equation 4 into equation 1, we get:
x + y + 3x = 100
4x + y = 100 (equation 5)
Now we have two equations:
x - 3z = 0 (equation 4)
4x + y = 100 (equation 5)
To solve these equations, we can use substitution or elimination method.
Let's solve by substitution:
From equation 4, we have: x = 3z
Substituting this into equation 5, we get:
4(3z) + y = 100
12z + y = 100 (equation 6)
We now have two equations:
12z + y = 100 (equation 6)
x = 3z (equation 4)
We can solve equation 6 for y:
y = 100 - 12z
Substituting this into equation 4, we get:
x = 3z
Substituting the values of x and y into equation 1, we get:
3z + (100 - 12z) + z = 100
4z + 100 - 12z = 100
-8z = 0
z = 0
If z = 0, then x = 3z = 3(0) = 0
Substituting z = 0 into equation 6, we get:
y = 100 - 12(0)
y = 100
Therefore, the solution is:
x = 0
y = 100
z = 0
There are no one rupee or five rupee coins in the bag, and there are 100 two rupee coins.
Let's assume the number of one rupee coins is x, the number of two rupee coins is y, and the number of five rupee coins is z.
From the problem, we know the following equations:
1. x + y + z = 100 (equation 1) - The total number of coins is 100.
2. 1x + 2y + 5z = 200 (equation 2) - The total value of the coins is 200 rupees.
We can solve this system of equations to find the values of x, y, and z.
Multiplying equation 1 by 2, we get:
2x + 2y + 2z = 200 (equation 3)
Subtracting equation 2 from equation 3, we get:
(2x + 2y + 2z) - (1x + 2y + 5z) = 200 - 200
x - 3z = 0 (equation 4)
Substituting equation 4 into equation 1, we get:
x + y + 3x = 100
4x + y = 100 (equation 5)
Now we have two equations:
x - 3z = 0 (equation 4)
4x + y = 100 (equation 5)
To solve these equations, we can use substitution or elimination method.
Let's solve by substitution:
From equation 4, we have: x = 3z
Substituting this into equation 5, we get:
4(3z) + y = 100
12z + y = 100 (equation 6)
We now have two equations:
12z + y = 100 (equation 6)
x = 3z (equation 4)
We can solve equation 6 for y:
y = 100 - 12z
Substituting this into equation 4, we get:
x = 3z
Substituting the values of x and y into equation 1, we get:
3z + (100 - 12z) + z = 100
4z + 100 - 12z = 100
-8z = 0
z = 0
If z = 0, then x = 3z = 3(0) = 0
Substituting z = 0 into equation 6, we get:
y = 100 - 12(0)
y = 100
Therefore, the solution is:
x = 0
y = 100
z = 0
There are no one rupee or five rupee coins in the bag, and there are 100 two rupee coins.
What is the probability of getting more than 5 when dice is trown- a)0.75
- b)0.16
- c)0.26
- d)0.2
Correct answer is option 'B'. Can you explain this answer?
What is the probability of getting more than 5 when dice is trown
a)
0.75
b)
0.16
c)
0.26
d)
0.2
| | Ayesha Joshi answered |
Formula used:
Probability of occurrence of the event:
P(E) = n(E)/n(S)
Where,
n(E) = Number of favorable outcome
n(S) = Number of possible outcome
Calculation:
Total no. of outcomes = 1, 2, 3, 4, 5 & 6
Number of possible outcomes n(S) = 6
Number of favorable outcomes n(E) = 1
Using the above formula
Probability of getting more than 5 = 1/6
∴ The correct answer is 0.16
Comprehension:Read the following information and answer the questions given below:A has 28 elements, B has 32 elements and (A U B) has 40 elements.How many elements does A ∩ B have?- a)12
- b)8
- c)10
- d)20
Correct answer is option 'D'. Can you explain this answer?
Comprehension:
Read the following information and answer the questions given below:
A has 28 elements, B has 32 elements and (A U B) has 40 elements.
How many elements does A ∩ B have?
a)
12
b)
8
c)
10
d)
20
| | Ayesha Joshi answered |
Formula used:
n(A U B) = n(A) + n(B) - n(A ∩ B)
Calculation:
n(A) = 28
n(B) = 32
n(A U B) = 40
n(A ∩ B) = 32 + 28 - 40 = 20
Three fair coins are tossed simultaneously. Find the probability of getting one head.- a)2/3
- b)1/6
- c)3/8
- d)1/8
Correct answer is option 'C'. Can you explain this answer?
Three fair coins are tossed simultaneously. Find the probability of getting one head.
a)
2/3
b)
1/6
c)
3/8
d)
1/8
| | Ayesha Joshi answered |
Given:
There are 3 fair coins.
Formula Used:
Probability of occurrence of an event = P(E) = Number of favorable outcomes/Number of possible outcomes = n(E)/n(S)
Calculation:
We know that:
When three fair coins are tossed simultaneously,
The numbers of possible outcomes are n(S)
n(S) = 23 = 8 = {HHH, HHT, HTH, THH, TTT, TTH, THT, HTT}
The number of favourable outcomes is
n(E) = {HTT, THT, TTH} = 3
Probability = 3/8
∴ The probability of getting one head = 3/8.
A dice is thrown randomly. What is the probability that the number shown on the dice is not divisible by 3?- a)1/3
- b)2/3
- c)1/4
- d)2/5
Correct answer is option 'B'. Can you explain this answer?
A dice is thrown randomly. What is the probability that the number shown on the dice is not divisible by 3?
a)
1/3
b)
2/3
c)
1/4
d)
2/5
| | Orion Classes answered |
Formula used:
Probability of an event = Number of favorable outcomes/Total outcomes
Calculation:
Numbers on dice not divisible by 3 are {1, 2, 4, 5}
⇒ Number of favorable outcomes = 4
Total possible outcomes = 6
∴ The probability that the number shown on the dice is not divisible by 3 is 4/6 = 2/3
The sample space of four coins tossed together is:- a)8
- b)64
- c)32
- d)16
Correct answer is option 'D'. Can you explain this answer?
The sample space of four coins tossed together is:
a)
8
b)
64
c)
32
d)
16
| | Ayesha Joshi answered |
Number of coins tossed = 4
∴ Sample space of four coins tossed = 24 = 16
Three coins are tossed simultaneously. Find the probability of getting exactly two heads.- a)5/8
- b)1/8
- c)1/2
- d)3/8
Correct answer is option 'D'. Can you explain this answer?
Three coins are tossed simultaneously. Find the probability of getting exactly two heads.
a)
5/8
b)
1/8
c)
1/2
d)
3/8
| | Orion Classes answered |
Given:
Three coins are tossed simultaneously.
Formula:
Probability = Number of favorable outcomes/ Total number of outcomes.
Calculation:
When three coins are tossed then the outcome will be any one of these combinations. (TTT, THT, TTH, THH. HTT, HHT, HTH, HHH).
So, the total number of outcomes is 8.
Now, for exactly two heads, the favorable outcome is (THH, HHT, HTH).
We can say that the total number of favorable outcomes is 3.
Again, from the formula
Probability = Number of favorable outcomes/Total number of outcomes
Probability = 3/8
∴ The probability of getting exactly two heads is 3/8.
The probability of Sita, Gita and Mita passing a test is 60%, 40% and 20% respectively. What is the probability that at Sita and Gita will pass the test and Mita will not?- a)38.4%
- b)60%
- c)4.8%
- d)19.2%
Correct answer is option 'D'. Can you explain this answer?
The probability of Sita, Gita and Mita passing a test is 60%, 40% and 20% respectively. What is the probability that at Sita and Gita will pass the test and Mita will not?
a)
38.4%
b)
60%
c)
4.8%
d)
19.2%
 | Hannah Thompson answered |
To find the probability that Sita and Gita will pass the test and Mita will not, we need to multiply the individual probabilities of Sita and Gita passing the test and the probability of Mita not passing the test.
Given:
Probability of Sita passing the test = 60% = 0.60
Probability of Gita passing the test = 40% = 0.40
Probability of Mita not passing the test = 100% - Probability of Mita passing the test = 100% - 20% = 80% = 0.80
To find the probability of Sita and Gita passing the test and Mita not passing, we multiply the individual probabilities:
Probability (Sita and Gita pass test and Mita does not pass) = Probability (Sita passes test) * Probability (Gita passes test) * Probability (Mita does not pass)
= 0.60 * 0.40 * 0.80
Calculating the above expression, we get:
= 0.192
Therefore, the probability that Sita and Gita will pass the test and Mita will not is 19.2% (option D).
Given:
Probability of Sita passing the test = 60% = 0.60
Probability of Gita passing the test = 40% = 0.40
Probability of Mita not passing the test = 100% - Probability of Mita passing the test = 100% - 20% = 80% = 0.80
To find the probability of Sita and Gita passing the test and Mita not passing, we multiply the individual probabilities:
Probability (Sita and Gita pass test and Mita does not pass) = Probability (Sita passes test) * Probability (Gita passes test) * Probability (Mita does not pass)
= 0.60 * 0.40 * 0.80
Calculating the above expression, we get:
= 0.192
Therefore, the probability that Sita and Gita will pass the test and Mita will not is 19.2% (option D).
In a room there are eight couples. Out of them if 4 people are selected at random, the probability that they may be couples is- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
In a room there are eight couples. Out of them if 4 people are selected at random, the probability that they may be couples is
a)
b)
c)
d)
| | Orion Classes answered |
Concept:
1) Combination: Selecting r objects from given n objects.
- The number of selections of r objects from the given n objects is denoted by nCr
2) Probability of an event happening = Number of ways it can happen/Total number of outcomes
Note: Use combinations if a problem calls for the number of ways of selecting objects.
Calculation:
Given:
In a room, there are eight couples.
⇒ Eight couples = 16 peoples
We have to select four peoples out of 16 peoples.
⇒ Total possible cases = 16C4
Now, we have to select four people- they may be couples
So, we have to select two couples from eight couples.
⇒ Favourable cases = 8C2
Hence Required Probability = 
Two dice are rolled simultaneously then the probability that they show different numbers is- a)5/6
- b)1/5
- c)1/3
- d)2/3
Correct answer is option 'A'. Can you explain this answer?
Two dice are rolled simultaneously then the probability that they show different numbers is
a)
5/6
b)
1/5
c)
1/3
d)
2/3
| | Orion Classes answered |
Given:
Two dices
Formula used:
Probability = No of desired outcomes/Total no of outcomes
Calculation:
We know that when 2 dice are thrown then total no of outcome will be 36.
According to question,
Outcomes with different numbers on dice = 30
Total no of outcomes = 36
Probability = 30/36
⇒ 5/6
∴ The correct answer is 5/6.
A bag contains only red, green and white balls. The probability of selecting a red ball from the bag at random is 1/3 and that of selecting a white ball at random is 1/2. If the bag contains 9 green balls, the total number of balls in the bag is:- a)45
- b)48
- c)42
- d)54
Correct answer is option 'D'. Can you explain this answer?
A bag contains only red, green and white balls. The probability of selecting a red ball from the bag at random is 1/3 and that of selecting a white ball at random is 1/2. If the bag contains 9 green balls, the total number of balls in the bag is:
a)
45
b)
48
c)
42
d)
54
| | Ayesha Joshi answered |
Given:
Probability of red ball = 1/3
Probability of white ball = 1/2
Formula Used:
Probability = (Number of successful outcomes/Total number of outcomes)
P(E) = (nE)/(nS), where nE = Number of events and nS = Number of sample space
Calculation:
Probability of getting green ball = 1 - (1/3 + 1/2)
⇒ 1 - 5/6 = 1/6
According to the question:
If one unit corresponds to 9 green balls then,
6 unit = 6 × 9 = 54
Total no. Of balls = 54
∴ The total number of balls in the bag is 54.
A and B are two events such that P(B) = 0.4 and P(A ∪ B) = 0.6 If A and B are independent, then P(A) is- a)1/2
- b)1/3
- c)2/3
- d)2/5
Correct answer is option 'B'. Can you explain this answer?
A and B are two events such that P(B) = 0.4 and P(A ∪ B) = 0.6 If A and B are independent, then P(A) is
a)
1/2
b)
1/3
c)
2/3
d)
2/5
| | Ayesha Joshi answered |
Concept:
Independent events:
Two events are independent if the incidence of one event does not affect the probability of the other event.
If A and B are two independent events, then P(A ∩ B) = P(A) × P(B)
Calculation:
Given: P(B) = 0.4 and P(A ∪ B) = 0.6
P(A ∪ B) = 0.6
⇒ P(A) + P(B) - P(A ∩ B) = 0.6
⇒ P(A) + P(B) - P(A) × P(B) = 0.6 (∵ A and B are independent events.)
⇒ P(B) + P(A) [1 - P(B)] = 0.6
⇒ 0.4 + P(A) [1 - 0.4] = 0.6
⇒ P(A) × 0.6 = 0.2
If a coin is tossed thrice, find the probability of getting one or two heads.- a)4/5
- b)5/8
- c)3/4
- d)6/4
Correct answer is option 'C'. Can you explain this answer?
If a coin is tossed thrice, find the probability of getting one or two heads.
a)
4/5
b)
5/8
c)
3/4
d)
6/4
| | Ayesha Joshi answered |
Concept:
P(A) = n(A)/n(S)
Where n(A) = No. of favourable cases for event A and n(S) = cardinality of sample space.
Solution:
If a coin is tossed thrice, possible outcomes are:
S = {HHH, HHT, HTH, THH, THT, TTH, HTT, TTT}
Probability of getting one or two heads:
A = {HHT, HTH, THH, THT, TTH, HTT}
P(A) = 6/8
= 3/4
The probability of passing an examination for A and B are 0.7 and 0.8 respectively. Find the probability that at least one of them pass the examination.- a)0.6
- b)0.64
- c)0.94
- d)0.9
Correct answer is option 'C'. Can you explain this answer?
The probability of passing an examination for A and B are 0.7 and 0.8 respectively. Find the probability that at least one of them pass the examination.
a)
0.6
b)
0.64
c)
0.94
d)
0.9
| | Ayesha Joshi answered |
Given:
The probability of passing an examination for A and B are 0.7 and 0.8 respectively.
Formula Used:
Probability = Favorable outcome/sample space
Calculation:
⇒ Probability that A fail in examination = 1 - 0.7 = 0.3
⇒ Probability that B fail in examination = 1 - 0.8 = 0.2
⇒ required probability = 0.7 × 0.8 + 0.8 × 0.3 + 0.7 × 0.2 = 0.56 + 0.24 + 0.14 = 0.94
∴ Required probability = 0.94
Alternate Method:
Alternate Method:
Probability that at least one of them pass the examination = 1 - probability that both fails the examination
⇒ required probability = 1 - (0.3 × 0.2) = 0.94
∴ Required probability = 0.94
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