Question

Beth and her husband have 8 good friends (inlcuding her two sisters) 5 of whom are men and 3 of whom are women. She won a radio contest with 6 national tickets. Her and her husband will definitely be attending the game but she would like to select friends to attend with her for the other tickets. She is having a hard time deciding, so she puts all 8 names in a hat and randomy draws the names.

a. Probability that she selects both of her sisters?

b. probability that she selects at least 2 men?

c. probability that she selects an equal number of men and women?

d. if she has to chang her methodology because two of the friends are married and wont attend seprately. What is the probability that she slects the married couple?

Answer 1

There are 5 men and 3 women

a)

since she selects both sisters

so rest of 2 people can be selected out of rest of 6 people in 6C2 ways =15

total ways =selecting 4 people out of 8 = 8C4 =70

so probability =15/70 =3/14

b)

P(atleast 2 men) =1-P(0 or 1 men)

P(zero men) means 0 men and 3 women but she has to select 4 people

so P(zero men ) =0

P( 1 men) = selecting 1 men out 5 and selecting 3 women out of 3

               =(5C1 * 3C3)/(70)

               =5/70

               =1/14

c)

P(equal number of men and women) means P(2 men and 2 women ) = selecting 2 men out of 5 and selecting 2 women out 3 women

                                                                                          =(5C2 * 3C2)/(8C4)

                                                                                          =(10*3)/70

                                                                                          =30/70

                                                                                          =3/7

d)

P(select married couple)

after selecting married couple,we have to select 2 from rest of 6 people which can be done in 6C2 ways =15

so probability =15/70

                              =3/14