Question

An Olympic archer misses the bull's-eye 13% of the time. Assume each shot is independent of the others. If she shoots 7 arrows, what is the probability of each of the results described in parts a through f below?

a) Her first miss comes on the seventh arrow. The probability is .
b) She misses the bull's-eye at least once. The probability is .
c) Her first miss comes on the second or third arrow. The probability is .
d) She misses the bull's-eye exactly 3 times. The probability is .
e) She misses the bull's-eye at least 3 times. The probability is .
f) She misses the bull's-eye at most 3 times. The probability is .

Answer 1

To calculate the probabilities, we use the binomial probability formula. The probability of the described outcomes are: a) 0.0019 b) 0.624 c) 0.248 d) 0.098 e) 0.148 f) 0.880

Step-by-step explanation:

To calculate the probabilities described in parts a through f, we need to use the binomial probability formula. The formula is:

`P(x) = ^nC_x * p^x * q^(^n^-^x^)`

Where:

• P(x) is the probability of having x successes
• n is the number of trials (in this case, the number of arrows)
• x is the number of successes
• p is the probability of success (missing the bull's-eye, in our case)
• q is the probability of failure (hitting the bull's-eye)

By substituting the given values into the formula, we can calculate each probability.

a) P(first miss on 7th arrow) =
`(0.13)^6 * (0.87)^1` = 0.0019

b) To find the probability of missing at least once, we can find the complement of hitting the bull's-eye on all shots: 1 - P(no misses) =
`1 - (0.87)^7` = 0.624

c) P(first miss on 2nd or 3rd arrow) = P(first miss on 2nd) + P(first miss on 3rd) =
`(0.13)^1 * (0.87)^1 + (0.13)^2 * (0.87)^0` = 0.236 + 0.012 = 0.248

d) P(miss exactly 3 times) =
`^nC_3 * p^3 * q^(^n^-^3^)` =
`^7C_3 * (0.13)^3 * (0.87)^4` = 0.098

e) To find the probability of missing at least 3 times, we can calculate the probabilities of missing 3, 4, 5, 6, and 7 times and add them:

P(miss >= 3) = P(miss 3) + P(miss 4) + P(miss 5) + P(miss 6) + P(miss 7) =
`^7C_3 * (0.13)^3 * (0.87)^4 + 7C4 * (0.13)^4 * (0.87)^3 + 7C5 * (0.13)^5 * (0.87)^2 + 7C6 * (0.13)^6 * (0.87)^1 + 7C7 * (0.13)^7 * (0.87)^0` = 0.148

f) To find the probability of missing at most 3 times, we can calculate the probabilities of missing 0, 1, 2, and 3 times and add them:

P(miss <= 3) = P(miss 0) + P(miss 1) + P(miss 2) + P(miss 3) =
`(0.87)^7 + ^7C_1 * (0.13)^1 * (0.87)^6 + ^7C_2 * (0.13)^2 * (0.87)^5 + ^7C_3 * (0.13)^3 * (0.87)^4` = 0.880