Find the indicated probabilities using the geometric distribution, the Poisson distribution, or the binomial 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) $P$ (the first pass he completes is thehrst or second pass) $=0.864$ (Round to three decimal places as needed.) (c) $P$ (he does not complete his first two passes) = (Round to three decimal places as needed.)

Step 1 ：Given that the football player completes a pass 63.1% of the time, we can denote the probability of success (completing a pass) as \(p_{success} = 0.631\) and the probability of failure (not completing a pass) as \(p_{failure} = 1 - p_{success} = 0.369\).

Step 2 ：For part (a), we are asked to find the probability that the first pass he completes is the second pass. This means the first pass must be a failure and the second pass must be a success. The probability of this happening is the product of the probability of failure on the first pass and the probability of success on the second pass, which is \(p_{a} = p_{failure} \times p_{success} = 0.233\).

Step 3 ：For part (b), we are asked to find the probability that the first pass he completes is the first or second pass. This 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 happening is the sum of the probability of success on the first pass and the product of the probability of failure on the first pass and the probability of success on the second pass, which is \(p_{b} = p_{success} + (p_{failure} \times p_{success}) = 0.864\).

Step 4 ：For part (c), we are asked to find the probability that he does not complete his first two passes. This means both the first and second passes must be failures. The probability of this happening is the product of the probability of failure on the first pass and the probability of failure on the second pass, which is \(p_{c} = p_{failure} \times p_{failure} = 0.136\).

Step 5 ：Final Answer: (a) The probability that the first pass he completes is the second pass is \(\boxed{0.233}\). (b) The probability that the first pass he completes is the first or second pass is \(\boxed{0.864}\). (c) The probability that he does not complete his first two passes is \(\boxed{0.136}\).