Question 3: In this exercise, we assume that, when a child is born, its gender a

Question

Question 3: In this exercise, we assume that, when a child is born, its gender and day of birth are uniformly random and independent of other children. Thus, for each G E {boy, girl) and each D E Sun, Mon, Tue, Wednes, Thurs, Fri, Satur the probability that a child has gender G and is born on a Dday is equal to 1/14. Anil Maheshwari2 has two children You are given that at least one of Anil's kids is a boy who was born on a Sunday. Determine the probability that Anil has two boys. Question 4: You are given a fair red die and a fair blue die. Each of these two dice has the letter a on one face, the letter b on two faces, and the letter c on three faces. You roll both dice uniformly at random and independently of each other. Define the events A-"at least one of the two rolls results in the letter b" aIl B-"both rolls result in the same letter" . Determine Pr(A), Pr(B), and Pr(A | B)

Explanation / Answer

3. Anil has 2 children. Probability of each child being a boy or girl is independent of the other child being a boy or girl.

We know that one of the kids is a boy. We need to find the probability that both the kids are boys.

So we need to find the probability that the second child is a boy.

The probability that a child is a boy or a girl = 1/2 (from the set G)

Hence the probability that Anil has two boys, given that one of his two kids is a boy is equal to 1/2 = 0.5

4. For each die, probability of getting an "a" = 1/6;

Probability of getting a "b" = 2/6

Probability of getting a "c" = 3/6

P(A) = P(at least one of the two dice has letter b) = 1 - P(none of the rolls has letter "b") = 1 - (4/6) x (4/6) = 1-16/36 = 20/36 = 5/9

P(B) = P(both dice have the same letter) = P(both have letter a) + P(both have letter b) + P(both have letter c) = (1/6) x (1/6) + (2/6) x (2/6) + (3/6) x (3/6) = 1/36+ 4/36+ 9/36 = 14/36

P(A/B) = P(A and B) / P(B) = (4/36) /(14/36) = 4/14 = 2/7

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