Five coins are tossed 3200 times. Using the Poisson distribution , the approximate probability of getting five heads 2 times is

Answer

To solve the problem of finding the approximate probability of getting five heads 2 times when tossing five coins 3200 times using the Poisson distribution, we can follow these steps:

### Step 1: Define the Problem
We are tossing 5 coins 3200 times. We want to find the probability of getting exactly 5 heads (which is the maximum possible when tossing 5 coins) exactly 2 times.

### Step 2: Determine the Probability of Getting 5 Heads in One Toss
When tossing 5 coins, the probability of getting heads for each coin is \( \frac{1}{2} \). Therefore, the probability of getting 5 heads in one toss is:
\[ P(\text{5 heads}) = \left(\frac{1}{2}\right)^5 = \frac{1}{32} \]

### Step 3: Calculate the Expected Number of Times to Get 5 Heads
Let \( n = 3200 \) (the number of trials), and \( p = \frac{1}{32} \) (the probability of success). The expected number of occurrences (mean) of getting 5 heads is:
\[ \lambda = n \cdot p = 3200 \cdot \frac{1}{32} = 100 \]

### Step 4: Use the Poisson Distribution
The Poisson distribution can be used to approximate the probability of getting exactly \( k \) occurrences (in this case, \( k = 2 \)) when the mean number of occurrences is \( \lambda \). The formula for the Poisson probability is:
\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
Substituting \( \lambda = 100 \) and \( k = 2 \):
\[ P(X = 2) = \frac{e^{-100} \cdot 100^2}{2!} \]

### Step 5: Calculate \( P(X = 2) \)
Calculating \( 2! \):
\[ 2! = 2 \]
Now substituting into the formula:
\[ P(X = 2) = \frac{e^{-100} \cdot 100^2}{2} = \frac{e^{-100} \cdot 10000}{2} = 5000 e^{-100} \]

### Conclusion
The approximate probability of getting five heads 2 times when tossing five coins 3200 times is:
\[ P(X = 2) \approx 5000 e^{-100} \]