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

Options

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

Answer

To find the probability of obtaining the number 5 exactly 4 times when a die is thrown 7 times, we can use the formula for the probability of success in a Bernoulli trial.

### Step-by-Step Solution:

1. **Identify the parameters**:
- Number of trials (n) = 7 (since the die is thrown 7 times)
- Number of successes (r) = 4 (we want to get the number 5 exactly 4 times)
- Probability of success (p) = 1/6 (the probability of rolling a 5 on a die)
- Probability of failure (q) = 5/6 (the probability of not rolling a 5)

2. **Use the Binomial Probability Formula**:
The probability of getting exactly r successes in n trials is given by the formula:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where \(\binom{n}{r}\) is the binomial coefficient.

3. **Calculate the binomial coefficient**:
\[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4! \cdot 3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

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

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

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

7. **Combine the results**:
\[ P(X = 4) = 35 \times \frac{1}{1296} \times \frac{125}{216} \]
\[ = 35 \times \frac{125}{1296 \times 216} \]

8. **Calculate the denominator**:
\[ 1296 \times 216 = 279936 \]

9. **Final calculation**:
\[ P(X = 4) = \frac{35 \times 125}{279936} = \frac{4375}{279936} \]

### Final Answer:
Thus, the probability of obtaining the number 5 exactly 4 times when a die is thrown 7 times is:
\[ P(X = 4) = \frac{4375}{279936} \]