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Statistics:

1.A. A family has 2 children. Assuming that the sex of a child is independent of another child, and that male and female children are equally likely, calculate the following probabilities. Give all answers rounded to two places past decimal What is the probability that both children are girls? What is the probability that at least one of the children is female? What is the probability that both children are girls if you know that there is at least one daughter? 1.B. Experience has shown that the probability of passing a particular exam if a student studies for 6 hours or more the previous day is 0.8, and that the probability of passing this exam is only 0.36 if the student studies for less than 6 hours the previous day. 72 percent of students actually do study for 6 or more hours the day before the exam. Suppose a student is chosen at random, and the number of hours they studied the previous day is not known. Answer using 3 digits past the decimal place What is the probability that the student studied for at least 6 hours and passes the exam? What is the probability that the student passes the exam? If the student passes the exam, what is the probability that they studied for 6 or more hours the previous day? 1.C The number of bus rides taken by Dalhousie students on a weekday is a random variable X with probability distribution given below: 0 2 0.2 P(x) What is the probability that at least one ride is taken. Answer to 2 digits past the decimal. What is the probability that at most one ride is taken? Answer to 2 digits past the decimal. Find the mean number of bus rides taken by Dalhousie students on weekdays. Answer to 2 digits past the decimal Find the variance of X. Answer to 2 digits past the decimal. 0.7 0.1

Explanation / Answer

1.A.

Let B shows the boy and G shows the girl. So possible sample space for 2 chidren is

S = {BB, BG, GB, GG}

These all 4 outcomes are equally likley. The probability that both chidren are girl is

P(GG) = 1/ 4= 0.25

The probability that at least one of the children is girl will be

P(at least one is girl) = P(BG) + P(GB) + P(GG) = 1/4 + 1/4 + 1/4 = 0.75

The probabilty that both are girls given that at least one is daughter is

P(both girls | at least one daughter) = P(both girls and at least one daughter) / P(at least one daughter) =  P(both girls) / P(at least one daughter) = 0.25 / 0.75 = 1/3 = 0.33

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