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← JEE Advanced 2021 · Paper 1
MathematicsNumerical+2Question 44 of 57

JEE Advanced 2021 · Paper 1, Mathematics question 6

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 1254p2\dfrac{125}{4}p_2 is ____.

Answer and worked solution

Official answer: 24.4

Complement. The minimum is at least 41 exactly when all three numbers are at least 41:
p2=1−(0.6)3=0.784.p_2=1-(0.6)^3=0.784.

Value.
1254×0.784=24.5.\frac{125}4\times0.784=24.5.