Question

An Olympic archer misses the bull's-eye 12% of the time. Assume each shot is independent of the others. If she shoots 9 arrows, what is the probability of each of the results described in parts a through f below?

Answer 1

To calculate the probabilities, we assume that each shot is independent. The probability of each of the given results can be found by multiplying the probabilities of the individual shots.

Step-by-step explanation:

The probability of missing the bull's-eye is given as 12%. Let's calculate the probability of each of the results described in parts a through f:

1. The probability of missing the bull's-eye on one shot is 12%. The probability of doing this on two consecutive shots (independent events) is 12% x 12% = 0.12 x 0.12 = 0.0144, or 1.44%.
2. The probability of hitting the bull's-eye on one shot is 100% - 12% = 88%. The probability of doing this on two consecutive shots is 88% x 88% = 0.88 x 0.88 = 0.7744, or 77.44%.
3. The probability of hitting the bull's-eye on one shot is 88%. The probability of missing it on the second shot is 12%. Assuming independence, the probability of this sequence of events is 88% x 12% = 0.88 x 0.12 = 0.1056, or 10.56%.
4. The probability of missing the bull's-eye on one shot is 12%. The probability of hitting it on the second shot is 88%. The probability of this sequence of events is 12% x 88% = 0.12 x 0.88 = 0.1056, or 10.56%.
5. The probability of hitting the bull's-eye on one shot is 88%. The probability of missing it on the second shot is 12%. The probability of this sequence of events is 88% x 12% = 0.88 x 0.12 = 0.1056, or 10.56%.
6. The probability of missing the bull's-eye on one shot is 12%. The probability of missing it on the second shot is also 12%. The probability of this sequence of events is 12% x 12% = 0.12 x 0.12 = 0.0144, or 1.44%.