Question

A certain weightlifter is prone to back injury. He finds that he has a 20% chance of hurting his back if he uses the proper form of bending at the hips and keeping his spine locked. The probability that he will hurt his back with bad form is 95%. The probability that he uses proper form is 75%.

A. What is the probability that he hurts his back and has proper form?

B. What is the probability that he doesn't hurt his back and has proper form?

C. What is the probability he doesn't hurt his back?

Answer 1

Let Probability of Hurting his back = H

Let probability of using Proper Form = P

Let probability of using Bad Form = B

By Bayes Theorem, P(A given B) = P(A/B) = P(A and B) / P(B)

Given:

P(H/PF) = 0.2

P(H/BF) = 0.95 and

P(PF) = 0.75, therefore P(BF) = 0.25

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(A) P(Hurts his back and has Proper form) = P(H and PF)

P(H/PF) = 0.2 = P(H and PF) / P(PF)

Therefore P(H and PF) = P(H/PF) * P(PF)

Therefore P(H and PF) = 0.2 * 0.75 = 0.15

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(B) P(H/PF) = 0.2, therefore P(H'/PF) = 1 - 0.2 = 0.8, where P(H') is the probability that he does not hurt his back.

Therefore P(H'/PF) = P(H' and PF) / P(PF)

Therefore P(H'/PF) P(H' and PF) = P(H'/PF) * P(PF) = 0.8 * 0.75 = 0.6

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(C) P(H) = P(H/PF) * P(PF) + P(H/BF) * P(BF)

P(H) = (0.2 * 0.75) + (0.95 * 0.25)

P(H) = 0.15 + 0.2375 = 0.3875

Therefore P(H') = P(Not hurting his back) = 1 - P(H) = 1 - 0.3875 = 0.6125

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