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- Question
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
- Correct Answer
- 1/3
ExplanationIn 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. - 1. 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
- 2. 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
- 3. 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
- 4. 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
- 5. 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 Discuss
- 6. If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is ?
Options- A. 1
- B. 1/2
- C. 0
- D. 3/5 Discuss
- 7. In a party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples ?
Options- A. 6/7
- B. 19/21
- C. 7/31
- D. 5/21 Discuss
- 8. In a simultaneous throw of two dice, what is the probability of getting a doublet ?
Options- A. 1/6
- B. 1/3
- C. 4/7
- D. 4/5 Discuss
- 9. A group of five persons is formed from five boys and four girls. Find the probability that there are at least two girls in the group?
Options- A. 110/129
- B. 5/6
- C. 121/126
- D. 3/7 Discuss
- 10. Two dice are rolled simultaneously. Find the probability of getting a total of 9.
Options- A. 1/3
- B. 1/9
- C. 8/9
- D. 9/10 Discuss
Probability problems
Search Results
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: 5:4
Explanation:
Here,P(A)=4/9=p/(p+q)
P(odd against event A)=q/p=5/4
Correct Answer: 112/495
Explanation:
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: 7/44
Explanation:
Here, n(E) =
And, n(S) =
P(S) =
= 7/44
Correct Answer: 1/2
Explanation:
The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2
P(E) = 1 - 1/2 = 1/2.
Correct Answer: 5/21
Explanation:
Number of ways of (selecting at least two couples among five people selected) = (?C? x ?C?)
As remaining person can be any one among three couples left.
Required probability = (?C? x ?C?)/¹?C?
= (10 x 6)/252 = 5/21
Correct Answer: 1/6
Explanation:
In a simultaneous throw of two dice, n(S) = 6 x 6 = 36
Let E = event of getting a doublet = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
Correct Answer: 110/129
Explanation:
Total number of possible ways =
Number of favorable cases =
Therefore, required probability = 105/126 = 5/6
Correct Answer: 1/9
Explanation:
S = { (1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (2, 1), (2, 2),.........(6, 5), (6, 6) }
=> n(S) = 6 x 6 = 36
E = {(6, 3), (5, 4), (4, 5), (3, 6) }
=> n(E) = 4
Therefore, P(E) = 4/36 = 1/9
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