The probability Michael makes a shot from his desk into the wastebasket is 0.6.
a) What is the probability Michael shoots and misses? P(Michael misses the shot) =
b) What is the probability Michael makes 3 shots in a row? P(Michael makes 3 shots in a row) =
c) What is the probability Michael misses his first three shots and makes his forth shot? P(no basket until the forth shot) =

a) P(Michael misses the shot) = 0.4

b) P(Michael makes 3 shots in a row) = 0.216

c) P(no basket until the fourth shot) =0.0384

Step-by-step explanation:

a) In this case there are two possible outcomes everytime attempts a shot, he either misses or makes the shot, so the sum of probabilities for these outcomes must be equal to 1. We know that the probability Michael makes the shot is 0.6, so:

P(Michael misses the shot) = 1 - P(Michael makes the shot)

P(Michael misses the shot) = 1 - 0.6 = 0.4

b) Since each attempt is independent from the last one, the probability he makes 3 shots in a row is the product of the probability he makes one shot multiplied by itself three times.

P(Michael makes 3 shots in a row) = P(Michael makes the shot)* P(Michael makes the shot)*P(Michael makes the shot)

P(Michael makes 3 shots in a row) =0.6*0.6*0.6 = 0.216

c) To solve this one we can follow the same rationale as the previous one.

P(no basket until the fourth shot) = P(Michael misses the shot)* P(Michael misses the shot)* P(Michael misses the shot)*P(Michael makes the shot)

P(no basket until the fourth shot) =0.4*0.4*0.4*0.6 = 0.0384

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step-by-step explanation:

y=sec x is the correct answer

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