Exam P Practice Problem 98 – flipping coins

Problem 98-A

Coin 1 is an unbiased coin, i.e. when flipping the coin, the probability of getting a head is 0.5. Coin 2 is a biased coin such that when flipping the coin, the probability of getting a head is 0.6. One of the coins is chosen at random. Then the chosen coin is tossed repeatedly until a head is obtained.

Suppose that the first head is observed in the fifth toss. Determine the probability that the chosen coin is Coin 2.

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      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.2856

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.3060

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.3295

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 0.3564

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 0.3690

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Problem 98-B

Box 1 contains 3 red balls and 1 white ball while Box 2 contains 2 red balls and 2 white balls. The two boxes are identical in appearance. One of the boxes is chosen at random. A ball is sampled from the chosen box with replacement until a white ball is obtained.

Determine the probability that the chosen box is Box 1 if the first white ball is observed on the 6th draw.

\text{ }

      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.7530

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.7632

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.7825

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 0.7863

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 0.7915

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probability exam P

actuarial exam

math

Daniel Ma

mathematics

geometric distribution

Bayes

Answers

\copyright 2017 – Dan Ma

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