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Updated on: 22 Jan 2020, 23:54
Originally posted by CareerGeek on 10 Jan 2020, 10:36.
Last edited by CareerGeek on 22 Jan 2020, 23:54, edited 1 time in total.
Last edited by CareerGeek on 22 Jan 2020, 23:54, edited 1 time in total.
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
31% (02:15) correct
69%
(01:35)
wrong
based on 85
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Adam and Bob shoot a target in succession until someone hits the target. The probability of Adam and Bob hitting the target is 0.25 and 0.20 respectively. Bob will shoot only if Adam misses the target. The probability that Bob hits the target is?
A. 0.15
B. 0.25
C. 0.375
D. 0.45
E. 0.75
Posted from my mobile device
A. 0.15
B. 0.25
C. 0.375
D. 0.45
E. 0.75
Posted from my mobile device
10 Jan 2020, 11:02
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Probability that Bob hits target in first chance= \(\frac{3}{4}*\frac{1}{5}= \frac{3}{20}\) = \(0.15\)
Probability that Bob hits target in second chance= \(\frac{3}{4}*\frac{4}{5} * \frac{3}{4} * \frac{1}{5}\) = \(\frac{3}{20} * \frac{3}{5}\)= \(0.15*0.6\)
Probability that Bob hits target in third chance= \(\frac{3}{4}*\frac{4}{5} * \frac{3}{4} * \frac{4}{5} * \frac{3}{4} * \frac{1}{5} = \frac{3}{20} * (\frac{3}{5})^2= 0.15*(0.6)^2\)
and so on
probability that Bob hits the target is= \(0.15 + 0.15*0.6 + 0.15*(0.6)^2+.....\) {sum of infinite GP}
= \(\frac{a}{1-r}\) = \(\frac{0.15}{1-0.6} = 0.375\)
Probability that Bob hits target in second chance= \(\frac{3}{4}*\frac{4}{5} * \frac{3}{4} * \frac{1}{5}\) = \(\frac{3}{20} * \frac{3}{5}\)= \(0.15*0.6\)
Probability that Bob hits target in third chance= \(\frac{3}{4}*\frac{4}{5} * \frac{3}{4} * \frac{4}{5} * \frac{3}{4} * \frac{1}{5} = \frac{3}{20} * (\frac{3}{5})^2= 0.15*(0.6)^2\)
and so on
probability that Bob hits the target is= \(0.15 + 0.15*0.6 + 0.15*(0.6)^2+.....\) {sum of infinite GP}
= \(\frac{a}{1-r}\) = \(\frac{0.15}{1-0.6} = 0.375\)
Dillesh4096
22 Jan 2020, 05:46
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nick1816
Since the question asks the probability of Bob hitting the target, isn't it understood that it's asking for the first attempt ?
