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a.) Mason is attempting to kill a fly with a fly swatter. He estimates that for

ID: 3073371 • Letter: A

Question

a.) Mason is attempting to kill a fly with a fly swatter. He estimates that for each attempt, he will be successful 20% of the time.Mason will attempt to kill the fly up to four times. After that, if he has not been successful, he will quit. Let X be the number of times mason is unsuccessful. Find the expected value for X.
b.) Jerry, Janice, and Joana each independently are tying to pass a course. It is known that jerry has a 60% of passing the course, Janice has a 70% chance of passing the course, and Joana has a 90% chance of passing the course. Let X be the number who pass he course. Find E(X). a.) Mason is attempting to kill a fly with a fly swatter. He estimates that for each attempt, he will be successful 20% of the time.Mason will attempt to kill the fly up to four times. After that, if he has not been successful, he will quit. Let X be the number of times mason is unsuccessful. Find the expected value for X.
b.) Jerry, Janice, and Joana each independently are tying to pass a course. It is known that jerry has a 60% of passing the course, Janice has a 70% chance of passing the course, and Joana has a 90% chance of passing the course. Let X be the number who pass he course. Find E(X).
b.) Jerry, Janice, and Joana each independently are tying to pass a course. It is known that jerry has a 60% of passing the course, Janice has a 70% chance of passing the course, and Joana has a 90% chance of passing the course. Let X be the number who pass he course. Find E(X).

Explanation / Answer

Solution:-

a) The expected value for X is 3.20.

P(Success) = 0.20

P(Failure) = 1 - 0.20

P(Failure) = 0.80

E(x) = n × p

E(x) = 4 × 0.80

E(x) = 3.20

b) E(x) = 2.2

P(Pass Jerry) = 0.60

P(Pass Janica) = 0.70

P(Pass Joana) = 0.90

E(x) = 1×0.60 + 1 × 0.70 + 1 × 0.90

E(x) = 0.60 + 0.70 + 0.90

E(x) = 2.2