Solve the problem.
-Four boys and three girls are seated, in a row, at random, to watch a play. What is the probability th
a) a girl is seated at each end of the row?
b) one of the boys is seated in the middle seat?
c) boys and girls are seated alternately?
A) a) $\frac { 1 } { 7 }$ b) $\frac { 4 } { 7 }$ c) $\frac { 1 } { 35 }$

B) a) $\frac { 3 } { 7 }$ b) $\frac { 4 } { 7 }$ c) $\frac { 1 } { 7 }$

C) a) $\frac { 2 } { 7 }$ b) $\frac { 1 } { 7 }$ c) $\frac { 1 } { 35 }$

D) a) $\frac { 1 } { 7 }$ b) $\frac { 3 } { 7 }$ c) $\frac { 1 } { 35 }$

Question

Solve the problem.
-Four boys and three girls are seated, in a row, at random, to watch a play. What is the probability th
a) a girl is seated at each end of the row?
b) one of the boys is seated in the middle seat?
c) boys and girls are seated alternately?
A) a) $\frac { 1 } { 7 }$ b) $\frac { 4 } { 7 }$ c) $\frac { 1 } { 35 }$

B) a) $\frac { 3 } { 7 }$ b) $\frac { 4 } { 7 }$ c) $\frac { 1 } { 7 }$

C) a) $\frac { 2 } { 7 }$ b) $\frac { 1 } { 7 }$ c) $\frac { 1 } { 35 }$

D) a) $\frac { 1 } { 7 }$ b) $\frac { 3 } { 7 }$ c) $\frac { 1 } { 35 }$

Quizplus · Accepted Answer

a) To have a girl at each end, we choose 2 girls from 3 (which is $ \binom{3}{2} = 3 $ ways), and then arrange the remaining 5 children (2 girls and 3 boys) in any order, which is $ 5! $ ways. The total number of ways to seat all 7 children is $ 7! $. So, the probability is $ \frac{3 \times 5!}{7!} = \frac{1}{7} $.b) There are 4 boys, so any one of them can sit in the middle seat. The remaining 6 positions can be filled by the other 6 children in any order. Thus, the probability is $ \frac{4 \times 6!}{7!} = \frac{4}{7} $.c) For boys and girls to sit alternately, starting with a boy, we have 2 arrangements: BGBGBGG or GBGBGBG. Since there are 4 boys and 3 girls, only the first arrangement is possible. We arrange the 4 boys in $ 4! $ ways and the 3 girls in $ 3! $ ways. Thus, the probability is $ \frac{4! \times 3!}{7!} = \frac{1}{35} $.

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