Question

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 68.4% 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. (a) p(the first pass he completes is the second pass) = (round to three decimal places as needed.) (b) p(the first pass he completes is the first or second pass) = (round to three decimal places as needed.) (c) p(he does not complete his first two passes) = (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.

Explanation:

Step1: Calculate \(P(\text{the first pass he completes is the second pass})\)

The probability of not - completing the first pass is \(1 - 0.684=0.316\), and the probability of completing the second pass is \(0.684\).
Using the formula for the probability of two independent events \(P(A\cap B)=P(A)\times P(B)\), where \(A\) is "not - complete the first pass" and \(B\) is "complete the second pass"
\(P(\text{the first pass he completes is the second pass})=(1 - 0.684)\times0.684\)
\(=0.316\times0.684 = 0.216\)

Step2: Calculate \(P(\text{the first pass he completes is the first or second pass})\)

Using the formula \(P(A\cup B)=P(A)+P(B)\) for mutually - exclusive events.
\(P(\text{complete first pass}) = 0.684\), \(P(\text{the first pass he completes is the second pass})=0.216\)
\(P(\text{the first pass he completes is the first or second pass})=0.684 + 0.216=0.9\)

Step3: Calculate \(P(\text{he does not complete his first two passes})\)

The probability of not - completing a pass is \(1 - 0.684 = 0.316\)
Since the passes are independent events, \(P(\text{not complete first two passes})=(1 - 0.684)\times(1 - 0.684)\)
\(=0.316\times0.316=0.100\)

Step4: Determine which events are unusual

An event is considered unusual if its probability \(P\leq0.05\)
For part (a), \(P = 0.216>0.05\)
For part (b), \(P = 0.9>0.05\)
For part (c), \(P=0.100>0.05\)

Answer:

(a) \(0.216\)
(b) \(0.9\)
(c) \(0.100\)
D. None of the events are unusual.