Solve the problem.
-Four boys and three girls are seated, in a row, at random, to watch a play. What is the probability that
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) $\frac { 3 } { 7 }$ b) $\frac { 4 } { 7 }$ c) $\frac { 1 } { 7 }$
B) a) $\frac { 1 } { 7 }$ b) $\frac { 4 } { 7 }$ c) $\frac { 1 } { 35 }$
C) a) $\frac { 2 } { 7 } \quad$ b) $\frac { 1 } { 7 } \quad$ 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 that
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) $\frac { 3 } { 7 }$ b) $\frac { 4 } { 7 }$ c) $\frac { 1 } { 7 }$
B) a) $\frac { 1 } { 7 }$ b) $\frac { 4 } { 7 }$ c) $\frac { 1 } { 35 }$
C) a) $\frac { 2 } { 7 } \quad$ b) $\frac { 1 } { 7 } \quad$ c) $\frac { 1 } { 35 }$
D) a) $\frac { 1 } { 7 }$ b) $\frac { 3 } { 7 }$ c) $\frac { 1 } { 35 }$

Quizplus · Accepted Answer

a) To find the probability that a girl is seated at each end, we first place the girls at the ends, which can be done in $3 \times 2 = 6$ ways (3 choices for the first girl and 2 for the second). The remaining 5 people (3 boys and 1 girl) can be seated in any order, which is $5!$ ways. The total number of ways to seat all 7 people without restriction is $7!$. So, the probability is $\frac{6 \times 5!}{7!} = \frac{1}{7}$.b) For a boy to be seated in the middle seat, there are 4 choices (any of the 4 boys). The remaining 6 people can be seated in any order, which is $6!$ ways. So, the probability is $\frac{4 \times 6!}{7!} = \frac{4}{7}$.c) For boys and girls to be seated alternately, starting with a boy, the arrangement must be BGBGBGG or GBGBGBG. There are $4!$ ways to arrange the boys and $3!$ ways to arrange the girls in each case, so $2 \times 4! \times 3!$ total arrangements. The probability is $\frac{2 \times 4! \times 3!}{7!} = \frac{1}{35}$.

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