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← JEE Advanced 2021 · Paper 1
MathematicsNumerical+2Question 43 of 57

JEE Advanced 2021 · Paper 1, Mathematics question 5

Passage

Question Stem for Question Nos. 5 and 6

Three numbers are chosen at random, one after another with replacement, from the set S={1,2,3,…,100}S = \{1, 2, 3, \ldots, 100\}. Let p1p_1 be the probability that the maximum of chosen numbers is at least 81 and p2p_2 be the probability that the minimum of chosen numbers is at most 40.

The value of 6254p1\dfrac{625}{4}p_1 is ____.

Answer and worked solution

Official answer: 76.1

Complement. The maximum is at most 80 exactly when all three numbers are at most 80:
p1=1−(0.8)3=0.488.p_1=1-(0.8)^3=0.488.

Value.
6254×0.488=76.25.\frac{625}4\times0.488=76.25.