JEE Advanced 2021 Paper 1Question 1
The value of $\frac{625}{4}p_1$ is ___ .
Probability
Answer$76.25$
Solution
$p_1=1-(0.8)^3=0.488$, so $\frac{625}{4}p_1=76.25$.Probability - JEE Mathematics question (JEE Advanced 2021 Paper 1)
Comprehension · Questions 1–1
Read the passage given below and answer the questions that follow.
Three numbers are chosen at random, one after another with replacement, from the set $S=\{1,2,3,\ldots,100\}$. Let $p_1$ be the probability that the maximum of chosen numbers is at least 81 and $p_2$ be the probability that the minimum of chosen numbers is at most 40.
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