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Probability question

2023 · 29 Jan · Shift 1 · Q26
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  5. /2023 · 29 Jan · Shift 1 · Q26

Probability question

2023 · 29 Jan · Shift 1 · Q26

JEE MainMathematicsProbabilityMCQ+4 / −1
Fifteen football players of a club-team are given 15 T-shirts with their names written on the backside. If the players pick up the T-shirts randomly, then the probability that at least 3 players pick the correct T-shirt is :
  1. A
    16\frac{1}{6}61​
  2. B
    215\frac{2}{15}152​
  3. C
    524\frac{5}{24}245​
  4. D
    0.08
View written solutionFree

Correct answer: D

  1. Interpretation

Each random assignment of 15 distinct T-shirts to 15 distinct players is a permutation of 15 objects.

We need the probability that at least 3 players get their own T-shirt, i.e. the number of fixed points in a random permutation is at least 3.


  1. Useful fact about fixed points

For a random permutation of nnn objects, the number of fixed points is approximately Poisson with mean 111 when nnn is reasonably large. Here n=15n=15n=15, and this approximation is very accurate.

So, P(X=r)≈e−11r!,P(X=r) \approx e^{-1}\frac{1}{r!},P(X=r)≈e−1r!1​, where XXX = number of players who pick the correct T-shirt.

We want P(X≥3)=1−P(X=0)−P(X=1)−P(X=2).P(X\ge 3)=1-P(X=0)-P(X=1)-P(X=2).P(X≥3)=1−P(X=0)−P(X=1)−P(X=2).

Using the Poisson(1)(1)(1) model, P(X=0)≈e−1,P(X=1)≈e−1,P(X=2)≈e−12.P(X=0)\approx e^{-1}, \quad P(X=1)\approx e^{-1}, \quad P(X=2)\approx \frac{e^{-1}}{2}. P(X=0)≈e−1,P(X=1)≈e−1,P(X=2)≈2e−1​.

Hence P(X≥3)≈1−e−1(1+1+12).P(X\ge 3)\approx 1-e^{-1}\left(1+1+\frac12\right).P(X≥3)≈1−e−1(1+1+21​).

Now, e−1≈0.3679,e^{-1}\approx 0.3679,e−1≈0.3679, so P(X≥3)≈1−0.3679×2.5=1−0.91975=0.08025.P(X\ge 3)\approx 1-0.3679\times 2.5=1-0.91975=0.08025.P(X≥3)≈1−0.3679×2.5=1−0.91975=0.08025.

Thus, P(X≥3)≈0.08.P(X\ge 3)\approx 0.08.P(X≥3)≈0.08.


  1. Match with the options

The closest and intended answer is:

0.08\boxed{0.08}0.08​

So the correct option is D.


  1. Verification with stored answer

Stored correct answer: D

Our derived answer: D

They agree.

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