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The probability of a man hitting a target is 0.25. He shoots seven times. What is the probability of him hitting the target at least twice?

Answer :

The probability of the man hitting the target at least twice out of 7 shots is approximately 0.7632, or 76.32%.

To find the probability of the man hitting the target at least twice out of 7 shots, we can use the binomial probability formula. The binomial probability formula calculates the probability of getting exactly k successes in n independent Bernoulli trials, where the probability of success in each trial is p.

The formula for the probability of exactly k successes in n trials is:

[tex]\[ P(X = k) = \binom{n}{k} \times p^k \times (1 - p)^{n - k} \][/tex]

Where:

[tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient (number of combinations of n items taken k at a time)

p is the probability of success (in this case, hitting the target)

n is the number of trials

k is the number of successes

To find the probability of hitting the target at least twice, we sum the probabilities of hitting the target exactly 2, 3, 4, 5, 6, and 7 times.

[tex]\[ P(\text{at least 2 successes}) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) \][/tex]

Using the given information:

p = 0.25 (probability of hitting the target)

n = 7 (number of shots)

Now, we can calculate each individual probability using the formula above and sum them up to find the total probability.

First, let's calculate each individual probability:

1. Probability of hitting the target exactly 2 times:

[tex]\[ P(X = 2) = \binom{7}{2} \times (0.25)^2 \times (1 - 0.25)^{7 - 2} \][/tex]

[tex]\[ P(X = 2) = \binom{7}{2} \times (0.25)^2 \times (0.75)^5 \][/tex]

Using the binomial coefficient formula [tex]\( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)[/tex], we have:

[tex]\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \][/tex]

Substituting the values:

[tex]\[ P(X = 2) = 21 \times (0.25)^2 \times (0.75)^5 \][/tex]

[tex]\[ P(X = 2) \approx 0.3115 \][/tex]

2. Probability of hitting the target exactly 3 times:

[tex]\[ P(X = 3) = \binom{7}{3} \times (0.25)^3 \times (0.75)^{7 - 3} \][/tex]

[tex]\[ P(X = 3) = \binom{7}{3} \times (0.25)^3 \times (0.75)^4 \][/tex]

Using the binomial coefficient formula, [tex]\( \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \)[/tex]

Substituting the values:

[tex]\[ P(X = 3) = 35 \times (0.25)^3 \times (0.75)^4 \][/tex]

[tex]\[ P(X = 3) \approx 0.2625 \][/tex]

Similarly, we can calculate probabilities for hitting the target exactly 4, 5, 6, and 7 times.

3. Probability of hitting the target exactly 4 times:

[tex]\[ P(X = 4) = \binom{7}{4} \times (0.25)^4 \times (0.75)^{7 - 4} \][/tex]

[tex]\[ P(X = 4) = \binom{7}{4} \times (0.25)^4 \times (0.75)^3 \][/tex]

Using the binomial coefficient formula, [tex]\( \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35 \)[/tex]

Substituting the values:

[tex]\[ P(X = 4) = 35 \times (0.25)^4 \times (0.75)^3 \][/tex]

[tex]\[ P(X = 4) \approx 0.1455 \][/tex]

4. Probability of hitting the target exactly 5 times:

[tex]\[ P(X = 5) = \binom{7}{5} \times (0.25)^5 \times (0.75)^{7 - 5} \][/tex]

[tex]\[ P(X = 5) = \binom{7}{5} \times (0.25)^5 \times (0.75)^2 \][/tex]

Using the binomial coefficient formula, [tex]\( \binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7 \times 6}{2 \times 1} = 21 \)[/tex]

Substituting the values:

[tex]\[ P(X = 5) = 21 \times (0.25)^5 \times (0.75)^2 \][/tex]

[tex]\[ P(X = 5) \approx 0.0422 \][/tex]

5. Probability of hitting the target exactly 6 times:

[tex]\[ P(X = 6) = \binom{7}{6} \times (0.25)^6 \times (0.75)^{7 - 6} \][/tex]

[tex]\[ P(X = 6) = \binom{7}{6} \times (0.25)^6 \times (0.75)^1 \][/tex]

Using the binomial coefficient formula, [tex]\( \binom{7}{6} = \frac{7!}{6!(7-6)!} = \frac{7}{1} = 7 \)[/tex]

Substituting the values:

[tex]\[ P(X = 6) = 7 \times (0.25)^6 \times (0.75) \][/tex]

[tex]\[ P(X = 6) \approx 0.0015 \][/tex]

6. Probability of hitting the target exactly 7 times:

[tex]\[ P(X = 7) = \binom{7}{7} \times (0.25)^7 \times (0.75)^{7 - 7} \][/tex]

[tex]\[ P(X = 7) = \binom{7}{7} \times (0.25)^7 \times (0.75)^0 \][/tex]

Using the binomial coefficient formula, [tex]\( \binom{7}{7} = 1 \)[/tex]

Substituting the values:

[tex]\[ P(X = 7) = 1 \times (0.25)^7 \times 1 \][/tex]

[tex]\[ P(X = 7) \approx 0.0000 \][/tex]

Now, let's sum up these probabilities to find the total probability of hitting the target at least twice:

[tex]\[ P(\text{at least 2 successes}) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) \][/tex]

[tex]\[ P(\text{at least 2 successes}) \approx 0.3115 + 0.2625 + 0.1455 + 0.0422 + 0.0015 + 0.0000 \][/tex]

[tex]\[ P(\text{at least 2 successes}) \approx 0.7632 \][/tex]

So, the probability of the man hitting the target at least twice out of 7 shots is approximately 0.7632, or 76.32%.

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Rewritten by : Jeany

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