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) $\frac { 1 } { 7 }$ b) $\frac { 3 } { 7 }$ c) $\frac { 1 } { 35 }$
B) a) $\frac { 1 } { 7 }$ b) $\frac { 4 } { 7 } \quad$ c) $\frac { 1 } { 35 }$
C) a) $\frac { 3 } { 7 }$ b) $\frac { 4 } { 7 }$ c) $\frac { 1 } { 7 }$
D) a) $\frac { 2 } { 7 }$ b) $\frac { 1 } { 7 } \quad$ c) $\frac { 1 } { 35 }$

Question

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) $\frac { 1 } { 7 }$ b) $\frac { 3 } { 7 }$ c) $\frac { 1 } { 35 }$
B) a) $\frac { 1 } { 7 }$ b) $\frac { 4 } { 7 } \quad$ c) $\frac { 1 } { 35 }$
C) a) $\frac { 3 } { 7 }$ b) $\frac { 4 } { 7 }$ c) $\frac { 1 } { 7 }$
D) a) $\frac { 2 } { 7 }$ b) $\frac { 1 } { 7 } \quad$ c) $\frac { 1 } { 35 }$

Quizplus · Accepted Answer

a) To have a girl at each end, we choose 2 girls out of 3 for the ends in $3C2 = 3$ ways, and the remaining 5 can be arranged in $5!$ ways. The total arrangements are $7!$, so the probability is $\frac{3 \times 5!}{7!} = \frac{1}{7}$.b) There are 4 boys, so any one of them can be in the middle seat. The probability is $\frac{4}{7}$ because there are 7 seats and 4 choices for a boy to be in the middle.c) For alternate seating, start with a boy or a girl. There are 2 ways to choose who starts (boy or girl). Then, arrange the boys (4!) and the girls (3!) in their spots. The total is $2 \times 4! \times 3!$, and the probability is $\frac{2 \times 4! \times 3!}{7!} = \frac{1}{35}$.

Four Boys and Three Girls Are Seated, in a Row 17\frac { 1 } { 7 }71​

Four Boys and Three Girls Are Seated, in a Row $\frac { 1 } { 7 }$