If a die is thrown 7 times, then the probability of obtaining 5 exactly 4 times is

Options

A. `""^7C_(4)(1/6)^4(5/6)^3`
B. `""^7C_(4)(1/6)^3(5/6)^4`
C. `(1/6)^4(5/6)^3`
D. `(1/6)^3(5/6)^4`

Answer

To solve the problem of finding the probability of obtaining the number 5 exactly 4 times when a die is thrown 7 times, we can use the binomial probability formula.

### Step-by-Step Solution:

1. **Identify the parameters:**
- Number of trials (n): 7 (since the die is thrown 7 times)
- Number of successes (k): 4 (we want to obtain a 5 exactly 4 times)
- Probability of success (p): \( \frac{1}{6} \) (the probability of rolling a 5 on a die)
- Probability of failure (q): \( 1 - p = \frac{5}{6} \) (the probability of not rolling a 5)

2. **Use the binomial probability formula:**
The binomial probability formula is given by:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
where \( \binom{n}{k} \) is the binomial coefficient.

3. **Calculate the binomial coefficient:**
\[ \binom{n}{k} = \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4! \cdot 3!} = \frac{7 \times 6}{2 \times 1} = 35 \]

4. **Calculate the probability:**
Substitute the values into the binomial formula:
\[ P(X = 4) = \binom{7}{4} \left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^{7-4} \]
\[ P(X = 4) = 35 \left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^3 \]

5. **Calculate \( \left(\frac{1}{6}\right)^4 \) and \( \left(\frac{5}{6}\right)^3 \):**
\[ \left(\frac{1}{6}\right)^4 = \frac{1}{1296} \]
\[ \left(\frac{5}{6}\right)^3 = \frac{125}{216} \]

6. **Combine the results:**
\[ P(X = 4) = 35 \cdot \frac{1}{1296} \cdot \frac{125}{216} \]
\[ = 35 \cdot \frac{125}{279936} \]
\[ = \frac{4375}{279936} \]

7. **Final Result:**
The probability of obtaining a 5 exactly 4 times when a die is thrown 7 times is:
\[ P(X = 4) = \frac{4375}{279936} \]