Find the indicated probabilities using the geometric distribution, the Poisson distribution, or the binomlal distribution. Then determine if the events are unusual. If convenient, use the appropriate probability table or technology to find the probabilities.
A football player completes a pass $63.1 \%$ 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.
(a) $\mathrm{P}$ (the first pass he completes is the second pass) $=0.233$
(Round to three decimal places as needed)
(b) $\mathrm{P}$ (the first pass he completes is the first or second pass)=
(Round to three decimal places as needed.)
Answer
Final Answer: \(\boxed{(b)\, The\, probability\, that\, the\, first\, pass\, he\, completes\, is\, the\, first\, or\, second\, pass\, is\, approximately\, 0.864.}\)
Step 1 ：The problem is asking for the probability of certain events given a known success rate for a football player completing a pass. This is a problem of geometric distribution, which is a probability distribution that describes the number of Bernoulli trials needed for a success to occur.
Step 2 ：For part (a), the first pass he completes is the second pass, which means the first pass is a failure and the second pass is a success. The probability of this event is the product of the probability of failure on the first pass and the probability of success on the second pass.
Step 3 ：For part (b), the first pass he completes is the first or second pass, which means either the first pass is a success, or the first pass is a failure and the second pass is a success. The probability of this event is the sum of the probabilities of these two scenarios.
Step 4 ：Let's denote the probability of success as \(p_{success} = 0.631\) and the probability of failure as \(p_{failure} = 0.369\).
Step 5 ：Then, the probability that the first pass he completes is the second pass is \(p_a = p_{failure} \times p_{success} = 0.232839\).
Step 6 ：And the probability that the first pass he completes is the first or second pass is \(p_b = p_{success} + p_{failure} \times p_{success} = 0.863839\).
Step 7 ：Final Answer: \(\boxed{(a)\, The\, probability\, that\, the\, first\, pass\, he\, completes\, is\, the\, second\, pass\, is\, approximately\, 0.233.}\)
Step 8 ：Final Answer: \(\boxed{(b)\, The\, probability\, that\, the\, first\, pass\, he\, completes\, is\, the\, first\, or\, second\, pass\, is\, approximately\, 0.864.}\)
