A person can hit a target with a probability of $\frac{5}{8}$ in a single shot. If he fires 10 shots, what is the probability that he will hit the target exactly twice?

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Accepted Answer

**Solution:**

1. Given:
   - Probability of hitting the target in a single shot, $p = \frac{5}{8}$
   - Probability of missing the target, $q = 1 - p = \frac{3}{8}$
   - Number of shots, $n = 10$
   - Number of hits required, $k = 2$

2. The binomial probability formula is:
   $$P(X = k) = C(n,k) 	imes p^k 	imes q^{n-k}$$

3. Calculate the number of combinations:
   $$C(10,2) = \frac{10!}{2! 	imes 8!} = \frac{10 	imes 9}{2 	imes 1} = 45$$

4. Substitute the values:
   $$P(X=2) = 45 	imes \left(\frac{5}{8}ight)^2 	imes \left(\frac{3}{8}ight)^8$$

5. Simplify the expression:
   $$= 45 	imes \frac{25}{64} 	imes \frac{3^8}{8^8} = 45 	imes \frac{25 	imes 3^8}{64 	imes 8^8}$$

6. Note that $64 = 8^2$, so:
   $$= 45 	imes \frac{25 	imes 3^8}{8^2 	imes 8^8} = 45 	imes \frac{25 	imes 3^8}{8^{10}}$$

7. Multiply 45 and 25:
   $$45 	imes 25 = 1125$$

8. Final probability:
   $$P = \frac{1125 	imes 3^8}{8^{10}}$$

**Answer:** $\boxed{\frac{1125 	imes 3^8}{8^{10}}}$

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