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21 Sep 2020, 05:15
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A student needs an alarm clock to wake up, which has proven to successfully wake him 80% of the time. If he hears the alarm in the morning the probability of writing the test is 0.9 and if he doesn't hear the alarm the probability is 0.5. If he doesn't write the test, what is the probability that he didn't hear the alarm clock?
A. 5/41
B. 4/9
C. 5/9
D. 36/41
E. 38/41
Are You Up For the Challenge: 700 Level Questions
A. 5/41
B. 4/9
C. 5/9
D. 36/41
E. 38/41
Are You Up For the Challenge: 700 Level Questions
21 Sep 2020, 07:34
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If we imagine 100 mornings, on 80 of them, he hears the alarm, and on 20 of them he does not. When he hears the alarm, 8 times he still doesn't take the test. When he does not hear the alarm, 10 times he doesn't take the test. So of the 18 times when he fails to take the test, on 10 of them he didn't hear the alarm, and the answer is 10/18 = 5/9.
General Discussion
21 Sep 2020, 05:53
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By Bayes Theorem, P(A given B) = P(A/B) = P(A and B) / P(B)
Let the Probability of writing the test = P(W)
To Find: P(Doesn't hear the Alarm given that he doesn't writes the test) = P(A'/W') = P(A' and W') / P(W')
(i) P(W) = P(A) * P(W/A) + P(A') * P(W/A')
(ii) P(A' and W') = 1 - P(A U B) = 1 - [P(A) + P(W) - P(A and W)
Data Given
Probability of hearing the Alarm = P(A) = 0.8
Therefore, Probability of not hearing the Alarm = P(A') = 1 - 0.8 = 0.2
Probability of writing the test given he hears the alarm = P(W/A) = 0.9
Therefore 0.9 = P(W and A) / 0.8
P(W and A) = 0.9 * 0.8 = 0.72
Probability of writing the test given he does not hear the alarm = P(W/A') = 0.5
(i) P(W) = P(A) * P(W/A) + P(A') * P(W/A') = (0.8 * 0.9) + (0.2 * 0.5) = 0.72 + 0.1 = 0.82
Therefore P(W') = 1 - 0.82 = 0.18
(ii) P(A' and W') = 1 - P(A or W) = 1 - [P(A) + P(W) - P(A and W)] = 1 - (0.8 + 0.82 - 0.72) = 0.1
Therefore, P(A'/W') = 0.1 / 0.18 = 10/18 = 5/9
Option C
Arun Kumar
Let the Probability of writing the test = P(W)
To Find: P(Doesn't hear the Alarm given that he doesn't writes the test) = P(A'/W') = P(A' and W') / P(W')
(i) P(W) = P(A) * P(W/A) + P(A') * P(W/A')
(ii) P(A' and W') = 1 - P(A U B) = 1 - [P(A) + P(W) - P(A and W)
Data Given
Probability of hearing the Alarm = P(A) = 0.8
Therefore, Probability of not hearing the Alarm = P(A') = 1 - 0.8 = 0.2
Probability of writing the test given he hears the alarm = P(W/A) = 0.9
Therefore 0.9 = P(W and A) / 0.8
P(W and A) = 0.9 * 0.8 = 0.72
Probability of writing the test given he does not hear the alarm = P(W/A') = 0.5
(i) P(W) = P(A) * P(W/A) + P(A') * P(W/A') = (0.8 * 0.9) + (0.2 * 0.5) = 0.72 + 0.1 = 0.82
Therefore P(W') = 1 - 0.82 = 0.18
(ii) P(A' and W') = 1 - P(A or W) = 1 - [P(A) + P(W) - P(A and W)] = 1 - (0.8 + 0.82 - 0.72) = 0.1
Therefore, P(A'/W') = 0.1 / 0.18 = 10/18 = 5/9
Option C
Arun Kumar
