Six coins are tossed `9600` times. The probability of getting `5` heads `2` times, by using poisson distribution, is

Options

A. `(300^(2).e^(-300))/(2)`
B. `(300.e^(-300))/(2)`
C. `300^(2).e^(-300)`
D. `300 e^(-2)`

Answer

To solve the problem of finding the probability of getting 5 heads 2 times when tossing 6 coins 9600 times using the Poisson distribution, we can follow these steps:

### Step 1: Determine the probability of getting 5 heads in a single toss of 6 coins.
When tossing 6 coins, the total number of outcomes is \(2^6 = 64\). The number of favorable outcomes for getting exactly 5 heads can be calculated using the combination formula \(\binom{n}{k}\), where \(n\) is the total number of coins and \(k\) is the number of heads.

\[ \text{Number of ways to get 5 heads} = \binom{6}{5} = 6 \]

Thus, the probability \(P\) of getting 5 heads in a single toss is:

\[ P(\text{5 heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{6}{64} = \frac{3}{32} \]

### Step 2: Calculate the expected number of times (λ) that we would get 5 heads in 9600 tosses.
Using the formula for the expected value in a Poisson distribution:

\[ \lambda = n \cdot P \]

where \(n\) is the number of trials (9600) and \(P\) is the probability of getting 5 heads.

\[ \lambda = 9600 \cdot \frac{3}{32} = 300 \]

### Step 3: Use the Poisson probability formula to find the probability of getting exactly 2 occurrences of 5 heads.
The Poisson probability formula is given by:

\[ P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \]

In this case, we want to find \(P(X = 2)\):

\[ P(X = 2) = \frac{e^{-300} \cdot 300^2}{2!} \]

Calculating \(2!\):

\[ 2! = 2 \cdot 1 = 2 \]

Thus, we can substitute back into the formula:

\[ P(X = 2) = \frac{e^{-300} \cdot 300^2}{2} \]

### Step 4: Final expression for the probability.
The final expression for the probability of getting exactly 5 heads 2 times in 9600 tosses is:

\[ P(X = 2) = \frac{e^{-300} \cdot 300^2}{2} \]

### Summary
The probability of getting 5 heads 2 times when tossing 6 coins 9600 times using the Poisson distribution is given by:

\[ P(X = 2) = \frac{e^{-300} \cdot 300^2}{2} \]