Question

the probability of hitting a target is 1/6. what is the probability of hitting the target at most twice in seven throws?

Explanation:

Step1: Identify the distribution

This is a binomial probability problem. Let \( X \) be the number of hits. \( X \sim B(n = 7, p=\frac{1}{6}) \), and we need \( P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2) \). The binomial probability formula is \( P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k} \), where \( C(n,k)=\frac{n!}{k!(n - k)!} \).

Step2: Calculate \( P(X = 0) \)

For \( k = 0 \), \( C(7,0)=\frac{7!}{0!(7 - 0)!}=1 \), \( p^{0}=1 \), \( (1 - p)^{7}=(\frac{5}{6})^{7} \). So \( P(X = 0)=1\times1\times(\frac{5}{6})^{7}\approx0.2791 \).

Step3: Calculate \( P(X = 1) \)

For \( k = 1 \), \( C(7,1)=\frac{7!}{1!(7 - 1)!}=7 \), \( p^{1}=\frac{1}{6} \), \( (1 - p)^{6}=(\frac{5}{6})^{6} \). So \( P(X = 1)=7\times\frac{1}{6}\times(\frac{5}{6})^{6}\approx7\times\frac{1}{6}\times0.3349\approx0.3907 \).

Step4: Calculate \( P(X = 2) \)

For \( k = 2 \), \( C(7,2)=\frac{7!}{2!(7 - 2)!}=\frac{7\times6}{2\times1}=21 \), \( p^{2}=(\frac{1}{6})^{2}=\frac{1}{36} \), \( (1 - p)^{5}=(\frac{5}{6})^{5} \). So \( P(X = 2)=21\times\frac{1}{36}\times(\frac{5}{6})^{5}\approx21\times\frac{1}{36}\times0.4019\approx0.2344 \).

Step5: Sum the probabilities

\( P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)\approx0.2791 + 0.3907+0.2344\approx0.9042 \).

Answer:

The probability is approximately \( 0.9042 \) (or the exact value can be calculated as \( (\frac{5}{6})^{7}+7\times\frac{1}{6}\times(\frac{5}{6})^{6}+21\times(\frac{1}{6})^{2}\times(\frac{5}{6})^{5}=\frac{5^{7}+7\times5^{6}\times1+21\times5^{5}\times1^{2}}{6^{7}}=\frac{78125 + 7\times15625+21\times3125}{279936}=\frac{78125+109375 + 65625}{279936}=\frac{253125}{279936}\approx0.9042 \)).