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A student needs an alarm clock to wake up, which has proven to success 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


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A student needs an alarm clock to wake up, which has proven to success 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.
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A student needs an alarm clock to wake up, which has proven to success 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
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A student needs an alarm clock to wake up, which has proven to success 21 Sep 2020, 13:16
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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

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We want to find a probability given a certain condition, this belongs to conditional probability and we need to use Bayes' theorem to find the formula. The formula we will use here is:

\(P(No Alarm, given No Test) = \frac{P(No Alarm & No Test) }{ P(No Test)}\)

P(No Alarm & No Test) = 20% of no Alarm * 50% of no test when no alarm = 10%.

By default, we have P(No Test) = 1 - P(Test) so let us find the probability there is a test to calculate P(No Test).

P(Test) = 80% alarm * 90% test under alarm + 20% * 50% = 72%+10% = 82%. Then P(No Test) = 18%.

Finally the answer is 10% / 18% = 5/9.

Ans: C
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A student needs an alarm clock to wake up, which has proven to success 22 Sep 2020, 00:58
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Out of 10 mornings, a student wakes up successfully 8 times and fail 2 times.

Alarm hear and test taken: 0.9.Therefore Alarm heard but test not taken: 100 - 0. 9 = 0.1

=> 8 * 0.1 = 0.8 times [Not taken test]

Alarm not heard : 0.5. Therefore alarm not heard and test not taken: 2 * 0.5 = 0.1

Total number of cases he failed to take the test: 0.8 + 0.1 = 0.9

Alarm not heard: 0.5

=> \(\frac{0.5 }{ 0.9}\) = \(\frac{5}{9}\)

Answer C
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A student needs an alarm clock to wake up, which has proven to success 22 Sep 2020, 13:56
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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
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(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)

From where you got these two formula and how P(w) is equals to P(W) = P(A) * P(W/A) + P(A') * P(W/A')

Can you please explain.
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A student needs an alarm clock to wake up, which has proven to success 22 Sep 2020, 18:22
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jrk23
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By Bayes Theorem, P(A given B) = P(A/B) = P(A and B) / P(B)



(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)

From where you got these two formula and how P(w) is equals to P(W) = P(A) * P(W/A) + P(A') * P(W/A')

Can you please explain.
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Hello jrk23.


(i) This is a derivation which is obtained using Bayes Theorem.


P(W) = P(writing the exam when the alarm rings) + P(Writing the exam, when the alarm does not ring)


P(W) = P(W and A) + P(W and A') (And means intersection \(\cap\)


By Bayes Theorem, P(W/A) = P(W and A) / P(A). Therefore \(P(A) * P(W/A) = P(A) * \frac{P(W \space and \space A)}{P(A)} = P(W \space and \space A)\).

Similarly \(P(A') * P(W/A') = P(A') * \frac{P(W \space and \space A')}{P(A')} = P(W \space and \space A')\).


Therefore P(W) = P(A) * P(W/A) + P(A') * P(W/A')


For part (ii), This is a derivation from Venn Diagrams. I've attached a diagram for your reference.

Hope This Helps

Arun Kumar
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