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- Question
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
- Correct Answer
- 3/4
ExplanationLet 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
- 1. A bag contains 6 white and 4 black balls .2 balls are drawn at random. Find the probability that they are of same colour.
Options- A. 1/2
- B. 7/15
- C. 8/15
- D. 1/9 Discuss
- 2. Two taps X and Y can fill a tank in 15 hours and 20 hours respectively. If the two taps are opened at 2 p.m., then at what time (in p.m.) should the tap X be closed to completely fill the tank at exactly 2 a.m.?
Options- A. 8
- B. 7
- C. 9
- D. 10 Discuss
- 3. A swimming pool is filled in 10 hours by three tankers A, B and C. The tanker C is twice as fast as B and B is twice as fast as A. How much time will tanker A alone take to fill the swimming pool?
Options- A. 90 hours
- B. 40 hours
- C. 70 hours
- D. 80 hours Discuss
- 4. Two pipes, when working one at a time, can fill a cistern in 3 hhrs and 4 hrs, resp while a third pipe can drain the cistern empty in 8 hrs. All the three pipes were opened together when the cistern was 1/12 full. How long did it take for the cistern to be completely full?
Options- A. 2 hrs
- B. 1 hr 45 min
- C. 2 hr 10 min
- D. 2 hr 11 min Discuss
- 5. An electric pump can fill a tank in 2 hours. Because of a leak in the tank, it was taking 21?2 hours to fill the tank. The leak can drain all the water off the tank in how many hours?
Options- A. 9
- B. 10
- C. 10.5
- D. 11.5 Discuss
- 6. 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
- 7. 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
- 8. 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
- 9. 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
- 10. 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 Discuss
Probability problems
Search Results
Correct Answer: 7/15
Explanation:
Let S be the sample space
Then n(S) = no of ways of drawing 2 balls out of (6+4) = 10 = =45
Let E = event of getting both balls of same colour
Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)
= = = 15+6 = 21
Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15
Correct Answer: 8
Correct Answer: 70 hours
Correct Answer: 2 hrs
Correct Answer: 10
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: 2/7
Correct Answer: 2/7
Explanation:
Total number of outcomes possible, n(S) = 10 + 25 = 35
Total number of prizes, n(E) = 10
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: 5/12
Explanation:
Clearly, 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.
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