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- Question
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
- Correct Answer
- 12
ExplanationTotal balls = 40
Red balls = 18
Let green balls are x
Then, (18/40) × (
- 1. 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
- 2. 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
- 3. A number X is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What is the probability that |X|<2
Options- A. 3/7
- B. 3/4
- C. 4/5
- D. 5/7 Discuss
- 4. In a simultaneous throw of pair of dice. Find the probability of getting the total more than 7.
Options- A. 1/2
- B. 5/12
- C. 7/15
- D. 3/12 Discuss
- 5. Two dice are rolled simultaneously. Find the probability of getting the sum of numbers on the on the two faces divisible by 3 or 4?
Options- A. 3/7
- B. 7/11
- C. 5/9
- D. 6/13 Discuss
- 6. 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
- 7. 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
- 8. 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
- 9. 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 Discuss
- 10. 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
Probability problems
Search Results
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: 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.
Correct Answer: 3/7
Explanation:
X can take 7 values.
To get |X|+2) take X={?1,0,1}
=> P(|X|<2) = Favourable CasesTotal Cases = 3/7
Correct Answer: 5/12
Explanation:
Here n(S) = (6 x 6) = 36
Let E = event of getting a total more than 7
= {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}
Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12.
Correct Answer: 5/9
Explanation:
Here n(S) = 6 x 6 = 36
E={(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3) ,(6,6),(1,3),(2,2),(2,6),(3,1),(3,5), (4,4),(5,3),(6,2)}
=> n(E)=20
Required Probability n(P) = n(E)/n(S) = 20/36 = 5/9.
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: 319/420
Explanation:
All the events are mutually exclusive hence,
Required probability = P(P)+P(Q)+P(R)+P(S)
=
Correct Answer: 24/625
Explanation:
Total no of ways = (14 ? 1)! = 13!
Number of favorable ways = (12 ? 1)! = 11!
So, required probability = = =
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
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
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