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- Question
Two dice are tossed. The probability that the total score is a prime number is:
Options- A. 5/12
- B. 1/6
- C. 1/2
- D. 7/9
- Correct Answer
- 5/12
ExplanationClearly, n(S) = (6 x 6) = 36.
Let E = Event that the sum is a prime number.
Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }
n(E) = 15.
P(E) = n(E)/n(S) = 15/36 = 5/12.
- 1. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
Options- A. 1/2
- B. 3/5
- C. 9/20
- D. 8/15 Discuss
- 2. In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
Options- A. 2/7
- B. 5/7
- C. 1/5
- D. 1/2 Discuss
- 3. A man and his wife appear in an interview for two vacancies in the same post. The probability of husband's selection is (1/7) and the probability of wife's selection is (1/5). What is the probability that only one of them is selected ?
Options- A. 2/7
- B. 1/7
- C. 3/4
- D. 4/5 Discuss
- 4. Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?
Options- A. 52/221
- B. 55/190
- C. 55/221
- D. 19/221 Discuss
- 5. A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?
Options- A. 1/4
- B. 1/2
- C. 3/4
- D. 7/12 Discuss
- 6. What is the probability of getting 53 Mondays in a leap year?
Options- A. 1/7
- B. 3/7
- C. 2/7
- D. 1 Discuss
- 7. A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
Options- A. 2/91
- B. 1/22
- C. 3/22
- D. 2/77 Discuss
- 8. Two dice are thrown together .What is the probability that the sum of the number on the two faces is divided by 4 or 6.
Options- A. 7/18
- B. 14/35
- C. 8/18
- D. 7/35 Discuss
- 9. A letter is takenout at random from 'ASSISTANT' and another is taken out from 'STATISTICS'. The probability that they are the same letter is :
Options- A. 35/96
- B. 19/90
- C. 19/96
- D. None of these Discuss
- 10. In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
Options- A. 21/46
- B. 1/5
- C. 3/25
- D. 1/50 Discuss
Probability problems
Search Results
Correct Answer: 9/20
Explanation:
Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P(E) = n(E)/n(S) = 9/20.
Correct Answer: 2/7
Explanation:
Total number of outcomes possible, n(S) = 10 + 25 = 35
Total number of prizes, n(E) = 10
Correct Answer: 2/7
Correct Answer: 55/221
Explanation:
We have n(s) = 52 = 52*51/2*1= 1326.
Let A = event of getting both black cards
B = event of getting both queens
A?B = event of getting queen of black cards
n(A) = = = 325, n(B)= = 4*3/2*1= 6 and n(A?B) = = 1
P(A) = n(A)/n(S) = 325/1326;
P(B) = n(B)/n(S) = 6/1326 and
P(A?B) = n(A?B)/n(S) = 1/1326
P(A?B) = P(A) + P(B) - P(A?B) = (325+6-1) / 1326 = 330/1326 = 55/221
Correct Answer: 3/4
Explanation:
Let A, B, C be the respective events of solving the problem and be the respective events of not solving the problem. Then A, B, C are independent event
are independent events
Now, P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4
P( none solves the problem) = P(not A) and (not B) and (not C)
=
=
=
=
Hence, P(the problem will be solved) = 1 - P(none solves the problem)
= = 3/4
Correct Answer: 2/7
Explanation:
1 year = 365 days . A leap year has 366 days
A year has 52 weeks. Hence there will be 52 Sundays for sure.
52 weeks = 52 x 7 = 364days
366 ? 364 = 2 days
In a leap year there will be 52 Sundays and 2 days will be left.
These 2 days can be:
1. Sunday, Monday
2. Monday, Tuesday
3. Tuesday, Wednesday
4. Wednesday, Thursday
5. Thursday, Friday
6. Friday, Saturday
7. Saturday, Sunday
Of these total 7 outcomes, the favourable outcomes are 2.
Hence the probability of getting 53 days = 2/7
Correct Answer: 2/91
Explanation:
Let S be the sample space.
Then, n(S) = number of ways of drawing 3 balls out of 15 = = = 455.
Let E = event of getting all the 3 red balls.
n(E) = = = 10.
=> P(E) = n(E)/n(S) = 10/455 = 2/91.
Correct Answer: 7/18
Explanation:
Clearly, n(S) = 6 x 6 = 36
Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.
Then,E = {(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)}
n(E) = 14.
Hence, P(E) = n(E)/n(S) = 14/36 = 7/18
Correct Answer: 19/90
Explanation:
Here N and C are not common and same letters can be A, I, S, T. Therefore
Probability of choosing A = = 1/45
Probability of choosing I = = 1/45
Probability of choosing S = = 1/10
Probability of choosing T = = 1/15
Hence, Required probability =
Correct Answer: 21/46
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