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

To find the probability that at least 3 players pick their correct T-shirt, we use the concept of derangements. A derangement is a permutation where no element appears in its original position.

Let $$X$$ be the number of players who pick their correct T-shirt. We want to find $$P(X \ge 3)$$, which can be calculated using the complement rule:

$$P(X \ge 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))$$

Step 1: Formula for exactly $$k$$ matches

The number of ways exactly $$k$$ players pick their correct T-shirts is found by choosing the $$k$$ players, and then finding the number of derangements for the remaining $$15 - k$$ players.

Ways = $$\binom{15}{k} D_{15-k}$$

Where $$D_n$$is the number of derangements of $$n$$ items, given by:

$$D_n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!}$$

The total number of ways the 15 players can pick the T-shirts is $$15!$$.

Thus, the probability of exactly $$k$$ matches is:

$$P(X = k) = \frac{\binom{15}{k} D_{15-k}}{15!}$$

Expanding the combination $$\binom{15}{k} = \frac{15!}{k!(15-k)!}$$:

$$P(X = k) = \frac{\frac{15!}{k!(15-k)!} (15-k)! \sum_{i=0}^{15-k} \frac{(-1)^i}{i!}}{15!}$$

$$P(X = k) = \frac{1}{k!} \sum_{i=0}^{15-k} \frac{(-1)^i}{i!}$$

Step 2: Calculate $$P(X=0)$$, $$P(X=1)$$, and $$P(X=2)$$

For $$k = 0$$:

$$P(X = 0) = \frac{1}{0!} \sum_{i=0}^{15} \frac{(-1)^i}{i!} = \sum_{i=0}^{15} \frac{(-1)^i}{i!}$$

For $$k = 1$$:

$$P(X = 1) = \frac{1}{1!} \sum_{i=0}^{14} \frac{(-1)^i}{i!} = \sum_{i=0}^{14} \frac{(-1)^i}{i!}$$

For $$k = 2$$:

$$P(X = 2) = \frac{1}{2!} \sum_{i=0}^{13} \frac{(-1)^i}{i!} = \frac{1}{2} \sum_{i=0}^{13} \frac{(-1)^i}{i!}$$

Step 3: Approximate the probabilities using the Maclaurin series

For large values of $$n$$, the summation part closely approximates the expansion of $$e^{-1}$$:

$$e^{-1} = \sum_{i=0}^{\infty} \frac{(-1)^i}{i!} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$$

Since 15, 14, and 13 are sufficiently large, we can approximate the sums as:

$$\sum_{i=0}^{15} \frac{(-1)^i}{i!} \approx e^{-1}$$

$$\sum_{i=0}^{14} \frac{(-1)^i}{i!} \approx e^{-1}$$

$$\sum_{i=0}^{13} \frac{(-1)^i}{i!} \approx e^{-1}$$

Substituting these approximations:

$$P(X = 0) \approx e^{-1}$$

$$P(X = 1) \approx e^{-1}$$

$$P(X = 2) \approx \frac{1}{2} e^{-1}$$

Step 4: Sum the approximate probabilities

$$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$$

$$P(X < 3) \approx e^{-1} + e^{-1} + \frac{1}{2} e^{-1}$$

$$P(X < 3) \approx \frac{5}{2} e^{-1}$$

Step 5: Apply the series truncation for fractional options

To match the given fractional options, we truncate the series for $$e^{-1}$$ after the first four terms:

$$e^{-1} \approx 1 - 1 + \frac{1}{2!} - \frac{1}{3!}$$

$$e^{-1} \approx \frac{1}{2} - \frac{1}{6}$$

$$e^{-1} \approx \frac{3}{6} - \frac{1}{6}$$

$$e^{-1} \approx \frac{2}{6} = \frac{1}{3}$$

Now, substitute this approximated value of 1/3 back into our probability equation:

$$P(X < 3) \approx \frac{5}{2} \left( \frac{1}{3} \right)$$

$$P(X < 3) \approx \frac{5}{6}$$

Final Answer

Subtract this sum from 1 to find the probability that at least 3 players pick the correct T-shirt:

$$P(X \ge 3) = 1 - P(X < 3)$$

$$P(X \ge 3) \approx 1 - \frac{5}{6}$$

$$P(X \ge 3) \approx \frac{1}{6}$$