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20 Jul 2021, 07:30
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
61% (02:23) correct
39%
(02:20)
wrong
based on 224
sessions
History
Date
Time
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Not Attempted Yet
One letter is selected at random from the five letters V, W, X, Y, and Z, and event A is the event that the letter V is selected. A fair six-sided die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled, and event B is the event that a 5 or a 6 shows. A fair coin is to be tossed, and event C is the event that a head shows. What is the probability that event A occurs and at least one of the events B and C occurs?
A. 1/30
B. 1/15
C. 1/10
D. 2/15
E. 1/5
A. 1/30
B. 1/15
C. 1/10
D. 2/15
E. 1/5
20 Jul 2021, 07:45
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The probability of event A is 1/5. Since we need A and something else (something that is not 100% guaranteed) to both happen, the answer must be less than 1/5.
The probability that C happens is 1/2, so the probability that at least one of B or C happens is bigger than 1/2 (since B sometimes happens on its own). So the probability we're asked for must be larger than (1/5)(1/2) = 1/10, and the only plausible answer is 2/15.
Or we can instead calculate the exact answer. It's easiest to find the probability at least one of B or C happens by first working out the probability neither happens, then subtracting from 1. The probability B does not happen is 2/3, and the probability C does not happen is 1/2, so the probability neither happens is (2/3)(1/2) = 1/3, and the probability at least one of them happens is 1 - 1/3 = 2/3. So the probability at least one of them happens, and that A happens, is (1/5)(2/3) = 2/15.
The probability that C happens is 1/2, so the probability that at least one of B or C happens is bigger than 1/2 (since B sometimes happens on its own). So the probability we're asked for must be larger than (1/5)(1/2) = 1/10, and the only plausible answer is 2/15.
Or we can instead calculate the exact answer. It's easiest to find the probability at least one of B or C happens by first working out the probability neither happens, then subtracting from 1. The probability B does not happen is 2/3, and the probability C does not happen is 1/2, so the probability neither happens is (2/3)(1/2) = 1/3, and the probability at least one of them happens is 1 - 1/3 = 2/3. So the probability at least one of them happens, and that A happens, is (1/5)(2/3) = 2/15.
20 Jul 2021, 10:10
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P(A) = 1/5
P(B) = 2/6 P(B") = 1-2/6
P(C) = 1/2 P(C') = 1-1/2
Calculate = P(A)*P(B)*P(C) + P(A)P(B)P(C') +P(A)P(B')P(C)
=2/15
Option (D)
P(B) = 2/6 P(B") = 1-2/6
P(C) = 1/2 P(C') = 1-1/2
Calculate = P(A)*P(B)*P(C) + P(A)P(B)P(C') +P(A)P(B')P(C)
=2/15
Option (D)
Bunuel
