Probability problems
- 1. A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
Options- A. 10/21
- B. 11/21
- C. 1/2
- D. 2/7 Discuss
Correct Answer: 10/21
Explanation:
Total number of balls = (2 + 3 + 2) = 7.
Let S be the sample space.
Then, n(S) = Number of ways of drawing 2 balls out of 7 = = 21
Let E = Event of drawing 2 balls, none of which is blue.
n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls = = 10
Therefore, P(E) = n(E)/n(S) = 10/ 21.
- 2. 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
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
- 3. 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 Also asked in: AIEEE, Bank Exams, CAT
Correct Answer: 12
Explanation:
Total balls = 40
Red balls = 18
Let green balls are x
Then, (18/40) × (
- 4. 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
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
- 5. 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
Correct Answer: 319/420
Explanation:
All the events are mutually exclusive hence,
Required probability = P(P)+P(Q)+P(R)+P(S)
=
- 6. 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 Also asked in: AIEEE, Bank Exams, CAT, GATE, Bank Clerk, Bank PO
Correct Answer: 24/625
Explanation:
Total no of ways = (14 ? 1)! = 13!
Number of favorable ways = (12 ? 1)! = 11!
So, required probability = = =
- 7. 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 Also asked in: AIEEE, Bank Exams, CAT, Bank Clerk, Bank PO
Correct Answer: 1/7
Explanation:
A 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
- 8. 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 Also asked in: AIEEE, Bank Exams, CAT, GATE, Bank Clerk, Bank PO
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
- 9. 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 Also asked in: AIEEE, Bank Exams, CAT, GATE, Analyst, Bank Clerk, Bank PO, Database Administrator, IT Trainer
Correct Answer: 1/6
Explanation:
Required probability is given by P(E) =
- 10. 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
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 =
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