Probability Question 337
Question: 3 friends A, B and C play the game ?Pahle hum pahle tum? in which they throw a die one after the other and the one who will get a composite number 1st will be announced as winner, if A started the game followed by B and then C then what is the ratio of their winning probabilities?
Options:
A) $ 9:6:4 $
B) $ 8:6:5 $
C) $ 10:5:4 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Probability of getting a composite number is $ 2/6=1/3 $
Probability that A will win the game is $ ( \frac{2}{3} )( \frac{1}{3} )+( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{1}{3} ) $
$ +( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{1}{3} )+…. $ $ =\frac{\frac{1}{3}}{1-\frac{8}{27}}=( \frac{1}{3}\times \frac{27}{19} )=\frac{9}{19} $
Probability that B will win the game is $ ( \frac{2}{3} )( \frac{1}{3} )+( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{1}{3} ) $ $ +( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{1}{3} )+… $ $ =\frac{\frac{2}{9}}{1-\frac{8}{27}}=( \frac{2}{9}\times \frac{27}{19} )=\frac{6}{19} $
Probability that C will win the game is $ ( \frac{2}{3} )( \frac{2}{3} )( \frac{1}{3} )+( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{1}{3} ) $ $ +( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{2}{3} )( \frac{1}{3} )+… $ $ =\frac{\frac{4}{27}}{1-\frac{8}{27}}=( \frac{4}{27}\times \frac{27}{19} )=\frac{4}{19} $
So required ratio is $ 9:6:4, $
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