JEE Advance - Mathematics (2021 - Paper 1 Online - No. 6)
Three numbers are chosen at random, one after another with replacement, from the set S = {1, 2, 3, ......, 100}. Let p1 be the probability that the maximum of chosen numbers is at least 81 and p2 be the probability that the minimum of chosen numbers is at most 40.
The value of $${{125} \over 4}{p_2}$$ is ___________.
The value of $${{125} \over 4}{p_2}$$ is ___________.
Answer
24.5
Explanation
p2 = 1 $$-$$ p (all three numbers are > 40)
$$ = 1 - {\left( {{{60} \over {100}}} \right)^3} = 1 - {{27} \over {125}} = {{98} \over {125}}$$
So, $${{125{p_2}} \over 4} = {{125} \over 4} \times {{98} \over {125}} = {{98} \over 4} = 24.50$$
$$ = 1 - {\left( {{{60} \over {100}}} \right)^3} = 1 - {{27} \over {125}} = {{98} \over {125}}$$
So, $${{125{p_2}} \over 4} = {{125} \over 4} \times {{98} \over {125}} = {{98} \over 4} = 24.50$$
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