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Questions › IIT JAM › IIT JAM 2021 › Q41
1 markStatistics › Probability and Bayes' theorem › Addition rule, complements and inclusion-exclusionModerateabout 2 min

IIT JAM 2021 Question 41: Probability and Bayes' theorem

Three independent rolls of a fair die determine winnings: Rs. 800 for three sixes, Rs. 400 for two sixes, Rs. 100 for exactly one six, and nothing otherwise. The probability that a gambler wins at least Rs. 100 is _______ (round off to 2 decimal places).

Answer and worked solution

Answer: 0.41 to 0.43

Winning at least Rs. 100 requires at least one six in the three rolls.

This probability is 1−(56)3=1−125216=91216=0.42131 - \left(\tfrac56\right)^3 = 1 - \tfrac{125}{216} = \tfrac{91}{216} = 0.42131−(65​)3=1−216125​=21691​=0.4213, within the accepted range. The answer is about 0.420.420.42.

Probability and Bayes' theorem has come up 190 times in 7 exams since 2004.

What this question tests

This question is on Addition rule, complements and inclusion-exclusion, part of Probability and Bayes' theorem in Statistics.

The probability of at least one of several events, and of none.

TestStop has 30 past-paper questions on this subtopic; the nearest in year are below.

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