Question

A football player completes a pass 63.4 ​% of the time. Find the probability that​ (a) the first pass he completes is the second​ pass, (b) the first pass he completes is the first or second​ pass, and​ (c) he does not complete his first two passes.

Answer 1

Let p denote the probability of "success" or the event that the football player completes a pass; and

q denote the probability of "failure" or the event that the football player does not complete a pass.

We are given that:

p = 0.634

q = 1 - p = 1 - 0.634 = 0.366

(a)

P(the first pass he completes is the second pass) = P(first pass results in a failure)*P(second pass results in a success)

= q*p

= 0.366*0.634

= 0.232044 [Answer]

(b)

P(the first pass he completes is the first or second pass) = P(first pass he completes is the first pass) + P(first pass he completes is the second pass)

= p + qp

= 0.634 + 0.366*0.634

= 0.866044 [Answer]

(c)

P(he does not complete his first two passes) = P(he does not complete his first pass)*P(he does not complete his second pass)

= q*q

= 0.366*0.366

= 0.133956 [Answer]

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