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Worked solution

Use the same equally likely four-person selections throughout

Four persons are chosen at random from ten persons consisting of four men and six women. Three of the women are sisters. Calculate the probability that the four persons chosen will

(i) consist of four women,

(ii) consist of two women and two men,

(iii) include the three sisters.

CJC · Discussion · Question 3

Answer

(i) 1/141/14; (ii) 3/73/7; (iii) 1/301/30.

Fix the common sample space

Every unordered selection of four distinct people is equally likely. There are (104)=210\binom{10}4=210 possible selections.

How did this part go?

(i) Select four women

There are (64)\binom64 favourable selections, so P(four women)=(64)/(104)=15/210=1/14P(\text{four women})=\binom64/\binom{10}4=15/210=1/14.

How did this part go?

(ii) Select two of each

Choose two men from four and two women from six: (42)(62)\binom42\binom62 favourable selections. Hence P(two men, two women)=(42)(62)/(104)=90/210=3/7P(\text{two men, two women})=\binom42\binom62/\binom{10}4=90/210=3/7.

How did this part go?

(iii) Include all three sisters

The three sisters are fixed members of the selection. Choose the fourth person from the other seven: (71)\binom71 favourable selections. Thus P(all three sisters)=(71)/(104)=7/210=1/30P(\text{all three sisters})=\binom71/\binom{10}4=7/210=1/30. That fourth person may be a man or another woman.

How did this part go?

Common mistakes

Including the three sisters leaves seven possible fourth people; it does not force an all-women group.

Solution notes

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