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- Question
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
- Correct Answer
- 2/13
ExplanationNumber of queen cards = 4
Number of ace cards = 4
P(either a queen or ace) = 4/52 + 4/52= 8/52 = 2/13
- 1. 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
- 2. 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
- 3. 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 Discuss
- 4. 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
- 5. 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
- 6. 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
- 7. 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
- 8. One card is drawn from deck of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the ball drawn is red.
Options- A. 1/3
- B. 1/2
- C. 1/4
- D. 1/6 Discuss
- 9. Find the probability that one ball is white when two balls are drawn at random from a basket that contains 9 red, 7 white and 4 black balls.
Options- A. 18/95
- B. 18/190
- C. 1/2
- D. 91/190 Discuss
- 10. A six-sided die with faces numbered 1 through 6 is rolled three times. What is the probability that the face with the number 6 on it will not face upward on all the three rolls?
Options- A. 215/216
- B. 1/216
- C. 215/256
- D. 1/52 Discuss
Probability problems
Search Results
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: 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: 5/36
Explanation:
The 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
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: 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
Correct Answer: 1/2
Explanation:
Total number of elementary events = 52
There are 26 red cards,out of which one red card can be drawn in ways =26.
So,required probability = 26/52 = 1/2
Correct Answer: 91/190
Explanation:
Total number of elementary events = ways =190
There are 7 white balls out of which one white can be drawn in 7 ways and one ball from remaining 13 balls can be drawn in ways.
So,favourable events =
Therfore,required probability = ( )/ =91/90
Correct Answer: 215/216
Explanation:
Probability of getting 6 at the top once =1/6
Probability of getting 6 at the top three times =1/6 x 1/6 x 1/6 =216
Probability of no getting 6 at he top any time = 1 - 1/216 = 215/216
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