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- Question
In a purse there are 30 coins, twenty one-rupee and remaining 50-paise coins. Eleven coins are picked simultaneously at random and are placed in a box. If a coin is now picked from the box, find the probability of it being a rupee coin?
Options- A. 4/7
- B. 2/3
- C. 1/2
- D. 5/6
- Correct Answer
- 2/3
ExplanationTotal coins 30
In that,
1 rupee coins 20
50 paise coins 10
Probability of total 1 rupee coins = 20C11Probability that 11 coins are picked = 30C11
Required probability of a coin now picked from the box is 1 rupee = 20C11/30C11 = 2/3.
- 1. In a box, there are 9 blue, 6 white and some black stones. A stone is randomly selected and the probability that the stone is black is ¼. Find the total number of stones in the box?
Options- A. 15
- B. 18
- C. 20
- D. 24 Discuss
- 2. The first 8 alphabets are written down at random. what is the probability that the letters b,c,d,e always come together ?
Options- A. 1/7
- B. 8!
- C. 7!
- D. 1/14 Discuss
- 3. fifteen persons are sitting around a circular table facing the centre. What is the probability that three particular persons sit together ?
Options- A. 3/91
- B. 2/73
- C. 1/91
- D. 3/73 Discuss
- 4. From a pack of 52 cards, 3 cards are drawn. What is the probability that one is ace, one is queen and one is jack?
Options- A. 19/5525
- B. 16/5525
- C. 17/5525
- D. 7/5525 Discuss
- 5. A room contains 3 brown, 5 black and 4 white chairs. Two chairs are picked and are put in the lawn. What is the probability that none of the chairs picked is white ?
Options- A. 14/33
- B. 14/55
- C. 12/55
- D. 13/33 Discuss
- 6. The probability of success of three students X, Y and Z in the one examination are 1/5, 1/4 and 1/3 respectively. Find the probability of success of at least two.
Options- A. 1/4
- B. 1/2
- C. 1/6
- D. 1/3 Discuss
- 7. I forgot the last digit of a 7-digit telephone number. If 1 randomly dial the final 3 digits after correctly dialing the first four, then what is the chance of dialing the correct number?
Options- A. 1/999
- B. 1/1001
- C. 1/1000
- D. 4/1000 Discuss
- 8. Out of 3 girls and 6 boys a group of three members is to be formed in such a way that at least one member is a girl. In how many different ways can it be done?
Options- A. 64
- B. 84
- C. 56
- D. 20 Discuss
- 9. In a single throw with 2 dices, what is probability of neither getting an even number on one and nor a multiple of 3 on other?
Options- A. 11/36
- B. 25/36
- C. 5/6
- D. 1/6 Discuss
- 10. If x is chosen at random from the set {1,2,3,4} and y is to be chosen at random from the set {5,6,7}, what is the probability that xy will be even?
Options- A. 1/2
- B. 2/3
- C. 3/4
- D. 4/3 Discuss
Probability problems
Search Results
Correct Answer: 20
Explanation:
We know that, Total probability = 1
Given probability of black stones = 1/4
=> Probability of blue and white stones = 1 - 1/4 = 3/4
But, given blue + white stones = 9 + 6 = 15
Hence,
3/4 ----- 15
1 ----- ?
=> 15 x 4/3 = 20.
Hence, total number of stones in the box = 20.
Correct Answer: 1/14
Explanation:
The 8 letters can be written in 8! ways.
n(S) = 8!
Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.
Now the letters a, (bcde), f, g and h can be arranged in 5! ways.
The letters b,c,d and e can be arranged themselves in 4! ways.
n(E) = 5! x 4!
Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14
Hence the answer is 1/14.
Correct Answer: 3/91
Explanation:
In a circle of n different persons, the total number of arrangements possible = (n - 1)!
Total number of arrangements = n(S) = (15 ? 1)! = 14 !
Taking three persons as a unit, total persons = 13 (in 4 units)
Therefore no. of ways for these 13 persons to around the circular table = (13 - 1)! = 12!
In any given unit, 3 particular person can sit in 3!. Hence total number of ways that any three person can sit =
n(E) = 12! X 3!
Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 12! X 3 ! / 14!
= 12!x3x2 / 14x13x12! = 6/14x13 = 3/91
Correct Answer: 16/5525
Explanation:
Required probability:
Correct Answer: 14/33
Explanation:
Total number of chairs = (3 + 5 + 4) = 12.
Let S be the sample space.
Then, n(s)= Number of ways of picking 2 chairs out of 12
= 12×11/2×1 = 66
Let n(E) = number of events of selecting 2 chairs for selecting no white chairs.
=> 8C2 = 8×7/2×1 = 28
Therefore required probability = 28/66 = 14/33.
Correct Answer: 1/6
Explanation:
P(X) = , P(Y) = , P(Z) =
Required probability:
= [ P(A)P(B){1?P(C)} ] + [ {1?P(A)}P(B)P(C) ] + [ P(A)P(C){1?P(B)} ] + P(A)P(B)P(C)
=
= = =
Correct Answer: 1/1000
Explanation:
It is given that last three digits are randomly dialled. then each of the digit can be selected out of 10 digits in 10 ways.
Hence required probability =
= 1/1000
Correct Answer: 64
Explanation:
Total number of possible ways = 9C3 = 84 ways
Required atleast one girl in the group of three = total possible ways - ways in which none is girl
None of the members in the group is girl = 6C3 = 20
Therefore, number of ways that at least one member is a girl in the group of three
= 84 - 20
= 64 ways.
Correct Answer: 25/36
Explanation:
We first calculate the probability of getting an even number on one and a multiple of 3 on other,Here, n(s) = 6x6 = 36 and
E = (2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4)(3,6) (6,2)(6,4)
n(E) = 11P(E) = 11/36Required probability = 1-11/36 = 25/36
Correct Answer: 2/3
Explanation:
S ={(1,5),(1,6),(1,7),(2,5),(2,6),(2,7),(3,5),(3,6),(3,7),(4,5),(4,6),(4,7)} xy will be even when even x or y or both will be even. P(E)=n(E)n(S)=812 = 4/3 |
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