# Shooters

In the army, regiments are six shooters. The first shooter target hit with a probability of 49%, next with 75%, 41%, 20%, 34%, 63%. Calculate the probability of target hit when shooting all at once.

### Correct answer:

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**Chris Roberts**

Let's denote:

H: Hit

M: Miss

Now, we can represent the probabilities as follows:

Shooter 1: P(H1) = 0.49, P(M1) = 1 - P(H1) = 0.51

Shooter 2: P(H2) = 0.75, P(M2) = 1 - P(H2) = 0.25

Shooter 3: P(H3) = 0.41, P(M3) = 1 - P(H3) = 0.59

Shooter 4: P(H4) = 0.20, P(M4) = 1 - P(H4) = 0.80

Shooter 5: P(H5) = 0.34, P(M5) = 1 - P(H5) = 0.66

Shooter 6: P(H6) = 0.63, P(M6) = 1 - P(H6) = 0.37

Now, we want to find the probability of all shooters hitting the target when shooting at once. We can use the multiplication rule for independent events, which states that the probability of all independent events occurring together is the product of their individual probabilities.

Let's calculate the probability of hitting the target when shooting all six shooters at once:

P(all shooters hit) = P(H1) * P(H2) * P(H3) * P(H4) * P(H5) * P(H6)

P(all shooters hit) = 0.49 * 0.75 * 0.41 * 0.20 * 0.34 * 0.63

P(all shooters hit) ≈ 0.008728 (rounded to six decimal places)

So, the probability of hitting the target when shooting all six shooters at once is approximately 0.008728 or about 0.873%.

H: Hit

M: Miss

Now, we can represent the probabilities as follows:

Shooter 1: P(H1) = 0.49, P(M1) = 1 - P(H1) = 0.51

Shooter 2: P(H2) = 0.75, P(M2) = 1 - P(H2) = 0.25

Shooter 3: P(H3) = 0.41, P(M3) = 1 - P(H3) = 0.59

Shooter 4: P(H4) = 0.20, P(M4) = 1 - P(H4) = 0.80

Shooter 5: P(H5) = 0.34, P(M5) = 1 - P(H5) = 0.66

Shooter 6: P(H6) = 0.63, P(M6) = 1 - P(H6) = 0.37

Now, we want to find the probability of all shooters hitting the target when shooting at once. We can use the multiplication rule for independent events, which states that the probability of all independent events occurring together is the product of their individual probabilities.

Let's calculate the probability of hitting the target when shooting all six shooters at once:

P(all shooters hit) = P(H1) * P(H2) * P(H3) * P(H4) * P(H5) * P(H6)

P(all shooters hit) = 0.49 * 0.75 * 0.41 * 0.20 * 0.34 * 0.63

P(all shooters hit) ≈ 0.008728 (rounded to six decimal places)

So, the probability of hitting the target when shooting all six shooters at once is approximately 0.008728 or about 0.873%.

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