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- Question
What is the probability that a leap year has 53 Saturdays and 52 Sundays ?
Options- A. 1/7
- B. 2/7
- C. 1/2
- D. 3/2
- Correct Answer
- 1/7
ExplanationA leap year has 52 weeks and two days
Total number of cases = 7
Number of favourable cases = 1
i.e., {Friday, Saturday}Required Probability = 1/7
- 1. 14 persons are seated around a circular table. Find the probability that 3 particular persons always seated together.
Options- A. 11/379
- B. 21/628
- C. 24/625
- D. 26/247 Discuss
- 2. In a race, the odd favour of cars P,Q,R,S are 1:3, 1:4, 1:5 and 1:6 respectively. Find the probability that one of them wins the race.
Options- A. 319/420
- B. 27/111
- C. 114/121
- D. 231/420 Discuss
- 3. There are four hotels in a town. If 3 men check into the hotels in a day then what is the probability that each checks into a different hotel?
Options- A. 1/2
- B. 3/4
- C. 4/7
- D. 3/8 Discuss
- 4. In a bag which contains 40 balls, there are 18 red balls and some green and blue balls. If two balls are picked up from the bag without replacement, then the probability of the first ball being red and second being green is 3/26. Find the number of blue balls in the bag.
Options- A. 16
- B. 12
- C. 10
- D. 14 Discuss
- 5. An unbiased die is tossed.Find the probability of getting a multiple of 3.
Options- A. 1/3
- B. 1/2
- C. 3/4
- D. 3/2 Discuss
- 6. A person starting with 64 rupees and making 6 bets, wins three times and loses three times, the wins and loses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is:
Options- A. A gain of Rs. 27
- B. A loss of Rs. 37
- C. A loss of Rs. 27
- D. A gain of Rs. 37 Discuss
- 7. Two letters are randomly chosen from the word TIME. Find the probability that the letters are T and M?
Options- A. 1/4
- B. 1/6
- C. 1/8
- D. 4 Discuss
- 8. An urn contains 4 white 6 black and 8 red balls . If 3 balls are drawn one by one without replacement, find the probability of getting all white balls.
Options- A. 5/204
- B. 1/204
- C. 13/204
- D. None of these Discuss
- 9. Two brother X and Y appeared for an exam. The probability of selection of X is 1/7 and that of B is 2/9. Find the probability that both of them are selected.
Options- A. 1/63
- B. 1/14
- C. 2/63
- D. 1/9 Discuss
- 10. Two cards are drawn from a pack of well shuffled cards. Find the probability that one is a club and other in King
Options- A. 1/52
- B. 1/26
- C. 1/13
- D. 1/2 Discuss
Probability problems
Search Results
Correct Answer: 24/625
Explanation:
Total no of ways = (14 ? 1)! = 13!
Number of favorable ways = (12 ? 1)! = 11!
So, required probability = = =
Correct Answer: 319/420
Explanation:
All the events are mutually exclusive hence,
Required probability = P(P)+P(Q)+P(R)+P(S)
=
Correct Answer: 3/8
Explanation:
Total cases of checking in the hotels = 4 x 4 x 4 = 64 ways.
Cases when 3 men are checking in different hotels = 4×3×2 = 24 ways.
Required probability =24/64 = 3/8
Correct Answer: 12
Explanation:
Total balls = 40
Red balls = 18
Let green balls are x
Then, (18/40) × (
Correct Answer: 1/3
Explanation:
Here S = {1,2,3,4,5,6}
Let E be the event of getting the multiple of 3
Then, E = {3,6}
P(E) = n(E)/n(S) = 2/6 = 1/3
Correct Answer: A loss of Rs. 37
Explanation:
As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the initial amount by 1.5 thrice and by 0.5 thrice (in any order).
The overall resultant will remain same.
So final amount with the person will be (in all cases):
= 64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs. 27
Hence the final result is:
64 ? 27 = 37
A loss of Rs.37
Correct Answer: 1/6
Explanation:
Required probability is given by P(E) =
Correct Answer: 1/204
Explanation:
Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then
Required probability =
=
Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9
When a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white
Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.
Hence the required probability =
Correct Answer: 2/63
Explanation:
Let A be the event that X is selected and B is the event that Y is selected.
P(A) = 1/7, P(B) = 2/9.
Let C be the event that both are selected.
P(C) = P(A) × P(B) as A and B are independent events:
= (1/7) × (2/9) = 2/63
Correct Answer: 1/26
Explanation:
Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
=P(A)+P(B)
=
= = =
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