Answer to Question #144820 in Statistics and Probability for Cameron Foster

Question

Three girls and three boys are randomly seated in a row. Please find the following. Keep you solutions
to 4 decimal places if needed .
(a) Calculate the Sample Space
(b) What is the probability if boys and girls alternate?
(c) What is the probabillity that boys sit together; girls sit together?
(d) What is the probability that girls sit at the end seats?
(e) What is the probability that Millium sits at the left end?

Answer 1

3 girls and 3 boys

a) Total number of boys and girls is 3+3=6. Thus sample space is 6!=720.

b) Boys and girls alternate: gbgbgb and bgbgbg.

The number of alternate arrangements is the number of possible arrangements of the girls multiplied by the number of possible arrangements of the boys: $2\cdot3!\cdot3!=72$

P(boys and girls alternate) = $\frac{2\cdot3!\cdot3!}{6!}=\frac{72}{720}=0.1$

c) Boys sit together; girls sit together: ggggbbb or bbbggg.

Let's consider the group of 3 boys as a single unit and the group of 3 girls as a single unit and arrange everyone accordingly.

The number of arranging 2 units a row is 2!=2.

The number of arranging boys among themselves is 3!=6.

The number of arranging girls among themselves is 3!=6.

Hence the probability is:

P(boys sit together and girls sit together) = $\frac{2\cdot3!\cdot3!}{6!}=\frac{72}{720}=0.1$

d) Girls sit at the end seats: bbbggg.

The number of arranging boys among themselves is 3!=6.

The number of arranging girls among themselves is 3!=6.

P(girls sit at the end seats) = $\frac{3!\cdot3!}{6!}=\frac{6\cdot6}{720}=0.05$

e) Millium sits at the left end:

Millium is at the left end so we are left with 5 places and 5 people.

P(Millium sits at the left end) = $\frac{5!}{6!}=\frac{120}{720}=0.1667$

Answer:

a) 720

b) 0.1

c) 0.1

d) 0.05

e) 0.1667