Questions: 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. (b) P (the first pass he completes is the first or second pass) = 864 (Round to three decimal places as needed.) (c) P (he does not complete his first two passes) = .136 (Round to three decimal places as needed.) Which of the events are unusual? Select all that apply. A. The event in part (a), P (the first pass he completes is the second pass), is unusual. B. The event in part (b), P (the first pass he completes is the first or second pass), is unusual. C. The event in part (c), P (he does not complete his first two passes), is unusual. D. None of the events are unusual.

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.
(b) P (the first pass he completes is the first or second pass) = 864
(Round to three decimal places as needed.)
(c) P (he does not complete his first two passes) = .136
(Round to three decimal places as needed.)
Which of the events are unusual? Select all that apply.
A. The event in part (a), P (the first pass he completes is the second pass), is unusual.
B. The event in part (b), P (the first pass he completes is the first or second pass), is unusual.
C. The event in part (c), P (he does not complete his first two passes), is unusual.
D. None of the events are unusual.

Transcript text: 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. (b) P (the first pass he completes is the first or second pass) $=864$ (Round to three decimal places as needed.) (c) P (he does not complete his first two passes) $=.136$ (Round to three decimal places as needed.) Which of the events are unusual? Select all that apply. A. The event in part (a), P (the first pass he completes is the second pass), is unusual. B. The event in part (b), $P$ (the first pass he completes is the first or second pass), is unusual. C. The event in part (c), P (he does not complete his first two passes), is unusual. D. None of the events are unusual.

Solution

Solution Steps

Step 1: Probability of Completing the Second Pass

To find the probability that the first pass he completes is the second pass, we use the formula for the geometric distribution:

\[ P(X = 2) = (1 - p)^{2-1} \cdot p \]

Substituting $ p = 0.631 $:

\[ P(X = 2) = (1 - 0.631)^{1} \cdot 0.631 = 0.369 \cdot 0.631 = 0.233 \]

Thus,

\[ \text{(a) } P(\text{the first pass he completes is the second pass}) = 0.233 \]

Step 2: Probability of Completing the First or Second Pass

Next, we calculate the probability that the first pass he completes is either the first or the second pass:

\[ P(X = 1) + P(X = 2) = p + P(X = 2) \]

Substituting the values:

\[ P(X = 1) = 0.631 \] \[ P(X = 2) = 0.233 \]

Thus,

\[ P(X = 1) + P(X = 2) = 0.631 + 0.233 = 0.864 \]

Therefore,

\[ \text{(b) } P(\text{the first pass he completes is the first or second pass}) = 0.864 \]

Step 3: Probability of Not Completing the First Two Passes

Now, we find the probability that he does not complete his first two passes:

\[ P(X > 2) = (1 - p)^2 \]

Substituting $ p = 0.631 $:

\[ P(X > 2) = (1 - 0.631)^2 = 0.369^2 = 0.136 \]

Thus,

\[ \text{(c) } P(\text{he does not complete his first two passes}) = 0.136 \]

Step 4: Determining Unusual Events

An event is considered unusual if its probability is less than $ 0.05 $. We evaluate each probability:

For part (a): $ 0.233 $ is not unusual.
For part (b): $ 0.864 $ is not unusual.
For part (c): $ 0.136 $ is not unusual.

Final Answer

\[ \text{(a) } P(\text{the first pass he completes is the second pass}) = 0.233 \] \[ \text{(b) } P(\text{the first pass he completes is the first or second pass}) = 0.864 \] \[ \text{(c) } P(\text{he does not complete his first two passes}) = 0.136 \] \[ \text{None of the events are unusual.} \]

Thus, the final boxed answers are:

\[ \boxed{(a) = 0.233, (b) = 0.864, (c) = 0.136} \]

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