Suppose Emerson loses 30% of all staring contests. (a) What is the probability that Emerson loses...

Suppose Emerson loses 30% of all staring contests.

(a) What is the probability that

Emerson loses two staring contests in a row?

(b) What is the probability that

Emerson loses three staring contests in a row?

(c) When events are independent, their complements are independent as well. Use this result to determine the probability that

Emerson loses three staring contests in a row, but does not lose four in a row.

Solutions

Expert Solution

(a)

Probability that Emerson loses staring contests = 0.3

Probability that Emerson loses two staring contests in a row = Probability that Emerson loses in first staring contest * Probability that Emerson loses in second staring contest

= 0.3 * 0.3 = 0.09

(b)

Probability that Emerson loses three staring contests in a row = Probability that Emerson loses in first staring contest * Probability that Emerson loses in second staring contest * Probability that Emerson loses in third staring contest

= 0.3 * 0.3 * 0.3 = 0.027

(c)

Probability that Emerson wins in first staring contest * Probability that Emerson loses in second staring contest * Probability that Emerson loses in third staring contest * Probability that Emerson loses in fourth staring contest

= 0.3 * 0.3 * 0.3 * (1 - 0.3) + (1 - 0.3) * 0.3 * 0.3 * 0.3

= 0.0189 + 0.0189

= 0.0378

orchestra answered 1 year ago

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