Question

1. The probability of a man hitting the target at a shooting range is .3. If...

1. The probability of a man hitting the target at a shooting range is .3. If he shoots 10 times, what is the probability that he hits the target exactly twice?

2. The probability of a man not hitting the target at a shooting range is .6. A success is defined as hitting the target. If he shoots 12 times, what is the probability that he misses the target just once?

3. The probability of a man not hitting the target at a shooting range is .6. A success is defined as hitting the target. If he shoots 12 times, what is the probability that he does hit the target at most 3 times?

4. P(A) = .6 and P(B) = .4. The events A and B have nothing in common. What is P(A and B)? ____

Please show how it is done. Have exam tomorrow and need to learn it. Thank you.

Homework Answers

Answer #1

1)

Here, n = 10, p = 0.3, (1 - p) = 0.7 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 2)
P(X = 2) = 10C2 * 0.3^2 * 0.7^8
P(X = 2) = 0.2335


2)


Here, n = 12, p = 0.6, (1 - p) = 0.4 and x = 1
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 1)
P(X = 1) = 12C1 * 0.6^1 * 0.4^11
P(X = 1) = 0.0003

3)

Here, n = 12, p = 0.4, (1 - p) = 0.6 and x = 3
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 3).
P(X <= 3) = (12C0 * 0.4^0 * 0.6^12) + (12C1 * 0.4^1 * 0.6^11) + (12C2 * 0.4^2 * 0.6^10) + (12C3 * 0.4^3 * 0.6^9)
P(X <= 3) = 0.0022 + 0.0174 + 0.0639 + 0.1419
P(X <= 3) = 0.2254


4)

P(A) = 0.6 , P (B) =0.4

P(A And B) = P(A) * P(B)
= 0.6 * 0.4
= 0.24

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