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Probability question

2021 · Shift 1 · Q25
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Probability question

2021 · Shift 1 · Q25

JEE AdvancedMathematicsProbabilityNumerical+2 / −1
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 1254p2{{125} \over 4}{p_2}4125​p2​ is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 24.5

  1. Total number of outcomes

Since 3 numbers are chosen with replacement from S={1,2,3,…,100},S=\{1,2,3,\dots,100\},S={1,2,3,…,100}, each choice has 100 possibilities.

So total outcomes are 1003.100^3.1003.

  1. Find p2p_2p2​: probability that the minimum is at most 40

We need p2=P(min⁡≤40).p_2=P(\min\le 40).p2​=P(min≤40).

It is easier to use the complement: P(min⁡>40).P(\min>40).P(min>40).

If the minimum is greater than 40, then all three numbers must be greater than 40.

Numbers greater than 40 in the set are 41,42,…,100,41,42,\dots,100,41,42,…,100, which are 100−40=60100-40=60100−40=60 numbers.

Thus, P(min⁡>40)=(60100)3=(35)3=27125.P(\min>40)=\left(\frac{60}{100}\right)^3=\left(\frac35\right)^3=\frac{27}{125}.P(min>40)=(10060​)3=(53​)3=12527​.

Therefore, p2=1−27125=98125.p_2=1-\frac{27}{125}=\frac{98}{125}.p2​=1−12527​=12598​.

  1. Compute 1254p2\dfrac{125}{4}p_24125​p2​

1254p2=1254⋅98125=984=492=24.5.\frac{125}{4}p_2=\frac{125}{4}\cdot \frac{98}{125}=\frac{98}{4}=\frac{49}{2}=24.5.4125​p2​=4125​⋅12598​=498​=249​=24.5.

  1. Final answer

The required value is 24.5.\boxed{24.5}.24.5​.

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