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21 Sep 2020, 05:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
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Question Stats:
50% (02:49) correct
50%
(03:00)
wrong
based on 72
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There are three sets of key rings A, B and C, for a house. The first set has five keys, the second has seven and the third has eight, of which only one key in each set opens the door to the store room. A keychain is chose at random followed by a key from the set. What is the probability that the chosen key opens the door and it came from the first key chain?
A. 131/840
B. 7/24
C. 56/131
D. 75/131
E. 17/24
Are You Up For the Challenge: 700 Level Questions
A. 131/840
B. 7/24
C. 56/131
D. 75/131
E. 17/24
Are You Up For the Challenge: 700 Level Questions
21 Sep 2020, 05:47
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Prob. of choosing right key from A = 1/3 * 1/5 = 1/15
Prob. of choosing right key from B = 1/7 * 1/3 = 1/21
Prob. of choosing right key from C = 1/8 * 1/3 = 1/24
Total Probability = 1/15 + 1/21 + 1/24 = 131/840. IMO A.
Prob. of choosing right key from B = 1/7 * 1/3 = 1/21
Prob. of choosing right key from C = 1/8 * 1/3 = 1/24
Total Probability = 1/15 + 1/21 + 1/24 = 131/840. IMO A.
21 Sep 2020, 06:38
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aritrar4
Hi Aritrar4. The question here is: What is the probability that the key came from the 1st Key Chain given that it opens the door.
By Bayes Theorem, P(A given B) = P(A/B) = \(\frac{P(A \space and \space B)}{P(B)}\)
Therefore P(Key Chain A / Opens) = \(\frac{P(Key \space Chain \space and \space Opens)}{P(Opens)}\)
P(Key Chain A and Opens) = \(\frac{1}{3} * \frac{1}{5} = \frac{1}{15}\)
The probability of P(Opens) = \(\frac{131}{840}\) (As found by you)
Therefore the required probability = \(\frac{\frac{1}{15}}{\frac{131}{840}} = \frac{1 \space * \space 840}{15 \space * \space 131} = \frac{56}{131}\)
Option C
Arun Kumar
