This question involves probability concepts, particularly binomial probability and complementary probability principles. Let’s solve it step by step.

Question Part (a): If he fires 5 times, what is the probability of hitting the target at least twice?

• The probability of hitting the target on a single trial is $p = \frac{1}{3}$, so the probability of missing is $q = 1 - p = \frac{2}{3}$.
• Number of trials is $n = 5$.
• "At least twice" means $X \geq 2$. We use the complement rule: $P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$

Step 1: Binomial formula

The probability of $X = k$ successes in $n$ trials is given by: $P(X = k) = \binom{n}{k} p^k q^{n-k}$

Step 2: Compute $P(X = 0)$ and $P(X = 1)$:

• For $X = 0$: $P(X = 0) = \binom{5}{0} \left(\frac{1}{3}\right)^0 \left(\frac{2}{3}\right)^5 = 1 \cdot 1 \cdot \left(\frac{32}{243}\right) = \frac{32}{243}$

• For $X = 1$: $P(X = 1) = \binom{5}{1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^4 = 5 \cdot \frac{1}{3} \cdot \frac{16}{81} = \frac{80}{243}$

Step 3: Compute $P(X \geq 2)$:

$P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$ $P(X \geq 2) = 1 - \frac{32}{243} - \frac{80}{243} = 1 - \frac{112}{243} = \frac{131}{243}$

Thus, the probability of hitting the target at least twice is: $P(X \geq 2) = \frac{131}{243} \approx 0.539 (53.9\%)$

Question Part (b): How many times should he fire so that the probability of hitting the target at least once is more than 90%?

• Probability of hitting at least once is $P(X \geq 1)$.
• Using the complement rule: $P(X \geq 1) = 1 - P(X = 0)$
• We require: $P(X \geq 1) > 0.9 \implies 1 - P(X = 0) > 0.9 \implies P(X = 0) < 0.1$

Step 1: Generalize $P(X = 0)$:

$P(X = 0) = \left(\frac{2}{3}\right)^n$ We need: $\left(\frac{2}{3}\right)^n < 0.1$

Step 2: Solve for $n$:

Taking the natural logarithm: $\ln\left(\frac{2}{3}\right)^n < \ln(0.1)$ $n \ln\left(\frac{2}{3}\right) < \ln(0.1)$ Since $\ln\left(\frac{2}{3}\right)$ is negative, divide by it and reverse the inequality: $n > \frac{\ln(0.1)}{\ln\left(\frac{2}{3}\right)}$ $n > \frac{-2.3026}{-0.4055} \approx 5.68$

Thus, he needs to fire at least 6 times to ensure the probability of hitting at least once exceeds 90%.

Final Answers:

(a) $P(X \geq 2) = \frac{131}{243} \approx 0.539$ (53.9%)
(b) He needs to fire at least 6 times.

Would you like a deeper explanation or have additional questions?
Here are 5 related questions to explore:

1. How is the binomial probability formula derived?
2. What happens to $P(X \geq 2)$ as the probability of hitting increases?
3. What is the impact of increasing the number of trials on $P(X \geq 2)$?
4. Can this question be solved using cumulative distribution tables for binomial probabilities?
5. How can this method be applied to real-world shooting scenarios?

Tip: Always use complement probabilities ($P(X \geq 1) = 1 - P(X = 0)$) to simplify calculations when "at least one" is involved.