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- Question
A fair six-sided die is rolled twice. What is the probability of getting 4 on the first roll and not getting 6 on the second roll ?
Options- A. 1/36
- B. 5/36
- C. 1/12
- D. 1/9
- Correct Answer
- 5/36
ExplanationThe two events mentioned are independent.
The first roll of the die is independent of the second roll. Therefore the probabilities can be directly multiplied.
P(getting first 4) = 1/6
P(no second 6) = 5/6
Therefore P(getting first 4 and no second 6) = 1/6 x 5/6 = 5/36
- 1. Let K and L be events on the same sample space, with P (K) = 0.8 and P (B) = 0.6. Are these two events are disjoint ?
Options- A. TRUE
- B. FALSE Discuss
- 2. What is the probability of getting a sum 9 from two throws of a dice ?
Options- A. 1/6
- B. 1/2
- C. 1/9
- D. 3/4 Discuss
- 3. Two dice are tossed. The probability that the total score is a prime number is :
Options- A. 1/6
- B. 1/2
- C. 5/12
- D. 3/4 Discuss
- 4. In a charity show tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random (blindly) will have a number with a hundredth digit of 2?
Options- A. 0.25
- B. 0.5
- C. 0.75
- D. 0.40 Discuss
- 5. You toss a coin AND roll a die. What is the probability of getting a tail and a 4 on the die?
Options- A. 1/2
- B. 1/12
- C. 2/3
- D. 3/4 Discuss
- 6. An unbiased die is rolled. Find the probability of getting a multiple of 3?
Options- A. 1/6
- B. 1/3
- C. 5/6
- D. None of these Discuss
- 7. The probabiltiy that throw of two dice yields a total of 5 or 6 is:
Options- A. 2/14
- B. 5/18
- C. 3/4
- D. 1/4 Discuss
- 8. The probability that a card drawn at random from the pack of playing cards may be either a queen or an ace is:
Options- A. 9/13
- B. 11/13
- C. 2/13
- D. None of these Discuss
- 9. What is the probability that a couple with four children have atleast one girl?
Options- A. 0.0625
- B. 0.9375
- C. 0.5
- D. 0.0257 Discuss
- 10. Find the probability of selecting exactly 2 children when four persons are choosen at random from a group of 3 men, 2 woman and 4 children.
Options- A. 9/29
- B. 10/21
- C. 12/21
- D. 14/19 Discuss
Probability problems
Search Results
Correct Answer: FALSE
Explanation:
These two events cannot be disjoint because P(K) + P(L) > 1.
P(A?B) = P(A) + P(B) - P(A?B).
An event is disjoint if P(A ? B) = 0. If K and L are disjoint P(K ? L) = 0.8 + 0.6 = 1.4
And Since probability cannot be greater than 1, these two mentioned events cannot be disjoint.
Correct Answer: 1/9
Explanation:
In two throws of a die, n(s)=(6 x 6)=36
let E= Event of geting a sum 9={(3,6),(4,5),(5,4),(6,3)}
P(E) = n(E)/n(S) = 4/36 = 1/9
Correct Answer: 5/12
Correct Answer: 0.40
Explanation:
250 numbers between 101 and 350 i.e. n(S)=250
n(E)=100th digits of 2 = 299?199 = 100
P(E)= n(E)/n(S) = 100/250 = 0.40
Correct Answer: 1/12
Explanation:
Probability of getting a tail when a single coin is tossed =12
Probability of getting 4 when a die is thrown =16
Required probability =(12)×(16)
= 1/12
Correct Answer: 1/3
Explanation:
S = { 1, 2, 3, 4, 5, 6 }
=> n(S) = 6
E = { 3, 6}
=> n(E) = 2
Therefore, P(E) = 2/6 =1/3
Correct Answer: 5/18
Explanation:
Total number of cases=36
Number of favourable cases(sum 5 or 6)=10
P(getting total of 5 or 6)=10/36=5/18
Correct Answer: 2/13
Explanation:
Number of queen cards = 4
Number of ace cards = 4
P(either a queen or ace) = 4/52 + 4/52= 8/52 = 2/13
Correct Answer: 0.9375
Explanation:
Here,n = 4(children)
P(girl)= 0.5
P(of atleast one girl)= 1 - P(no girls)
= 1 - 0.0625 = 0.9375
Correct Answer: 10/21
Explanation:
4 persons can be selected from 9 in ways =126
Fvaourable events = =60
So,required probability = 60/126 = 10/21
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