Discussion
Home ‣ Aptitude ‣ Probability Comments
- Question
A bag contains 7 green and 5 black balls. Three balls are drawn one after the other. The probability of all three balls being green, if the balls drawn are not replaced will be:
Options- A. 123/897
- B. 23/67
- C. 7/44
- D. 12/45
- Correct Answer
- 7/44
ExplanationHere, n(E) =
And, n(S) =
P(S) = = 7/44 - 1. A brother and a sister appear for an interview against two vacant posts in an office. The probability of the brother?s selection is 1/5 and that of the sister?s selection is 1/3. What is the probability that only one of them is selected?
Options- A. 1/5
- B. 3/4
- C. 2/5
- D. 3/5 Discuss
- 2. Out of first 20 natural numbers, one number is selected at random. The probability that it is either an even number or a prime number is ?
Options- A. 16/19
- B. 1
- C. 3/2
- D. 17/20 Discuss
- 3. A card is drawn from a pack of 52 cards. The probability of getting a queen of the club or a king of the heart is?
Options- A. 1/26
- B. 1/13
- C. 2/13
- D. 1/52 Discuss
- 4. In a simultaneous throw of two dice, what is the probability of getting a total of 7 ?
Options- A. 1/6
- B. 1/4
- C. 1/20
- D. 3/4 Discuss
- 5. Out of sixty students, there are 14 who are taking Economics and 29 who are taking Calculus. What is the probability that a randomly chosen student from this group is taking only the Calculus class ?
Options- A. 8/15
- B. 7/15
- C. 1/15
- D. 4/15 Discuss
- 6. One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn find the probability the card drawn is red.
Options- A. 2/3
- B. 1/2
- C. 1/52
- D. 13/51 Discuss
- 7. A bag contains 8 red and 4 green balls. Find the probability that two balls are red and one ball is green when three balls are drawn at random.
Options- A. 56/99
- B. 112/495
- C. 78/495
- D. None of these Discuss
- 8. If P(A)=4/9;then the odd against the event A is:
Options- A. 4:9
- B. 4:5
- C. 5:4
- D. 4:14 Discuss
- 9. If a card is drawn at random from a pack of 52 cards,what is the chance of getting a spade or ace?
Options- A. 0.25
- B. 5/13
- C. 0.20
- D. 4/13 Discuss
- 10. One lady has 2 children, one of her child is boy, what is the probability of having both are boys ?
Options- A. 1/3
- B. 1/2
- C. 2/3
- D. 2/5 Discuss
Probability problems
Search Results
Correct Answer: 2/5
Explanation:
Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected) + (Prob. that brother is not selected) × (Prob. that sister is selected)
= =
Correct Answer: 17/20
Explanation:
n(S) = 20
n(Even no) = 10 = n(E)
n(Prime no) = 8 = n(P)
P(E U P) = 10/20 + 8/20 - 1/20 = 17/20
Correct Answer: 1/26
Explanation:
Here in this pack of cards, n(S) = 52
Let E = event of getting a queen of the club or a king of the heart
Then, n(E) = 2
P(E) = n(E)/n(S) = 2/52 = 1/26
Correct Answer: 1/6
Correct Answer: 7/15
Explanation:
Given total students in the class = 60
Students who are taking Economics = 24 and
Students who are taking Calculus = 32
Students who are taking both subjects = 60-(24 + 32) = 60 - 56 = 4
Students who are taking calculus only = 32 - 4 = 28
probability that a randomly chosen student from this group is taking only the Calculus class = 28/60 = 7/15.
Correct Answer: 1/2
Explanation:
We know that:
When one card is drawn from a pack of 52 cards
The numbers of possible outcomes n(s) = 52
We know that there are 26 red cards in the pack of 52 cards
? The numbers of favorable outcomes n(E) = 26
Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S)
? required probability = 26/52 = 1/2.
Correct Answer: 112/495
Explanation:
Correct Answer: 5:4
Explanation:
Here,P(A)=4/9=p/(p+q)
P(odd against event A)=q/p=5/4
Correct Answer: 4/13
Explanation:
Number of spades in a standard deck of cards=13
Number of aces in a standard deck of cards=4
And,one of the aces is a spade.
So, 13 + 4 - 1 = 16 spades or aces to choose from.
Therfore,probabiltiy of getting a spade or an ace=16/52=4/13
Correct Answer: 1/3
Explanation:
In a family with 2 children there are four possibilities:
1) the first child is a boy and the second child is a boy (bb)
2) the first child is a boy and the second child is a girl (bg)
3) the first child is a girl and the second child is a boy (gb)
4) the first child is a girl and the second child is a girl (gg)
But already given that one child is boy. So we have three possibilities of (bb)(bg)(gb).
n(E)= both are boys=BB=1
n(S)= 3
Required probability P = n(E)/n(S) = 1/3.
Comments
There are no comments.More in Aptitude:
Programming
Copyright ©CuriousTab. All rights reserved.