JEE Advanced 2021 Paper 1 · Q05 · Probability

Correct answer:76.25

Solution

Three numbers chosen with replacement from $\{1,2,\ldots,100\}$. $p_1=P(\max\ge 81)=1-P(\text{all three}\le 80)=1-\left(\dfrac{80}{100}\right)^3=1-\dfrac{64}{125}=\dfrac{61}{125}$. Therefore $\dfrac{625}{4}p_1=\dfrac{625}{4}\cdot\dfrac{61}{125}=\dfrac{5\cdot 61}{4}=\dfrac{305}{4}=76.25$.

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