JEE Advanced 2021 Paper 1 · Q06 · Probability
The value of $\dfrac{125}{4} p_2$ is _____ .
Reveal answer + step-by-step solution
Correct answer:24.5
Solution
$p_2=P(\min\le 40)=1-P(\text{all three}>40)=1-\left(\dfrac{60}{100}\right)^3=1-\dfrac{27}{125}=\dfrac{98}{125}$. So $\dfrac{125}{4}p_2=\dfrac{125}{4}\cdot\dfrac{98}{125}=\dfrac{98}{4}=24.5$.
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