Question

4 boys and 2 girls occupy seats in a row at random. Then the probability that the girls occupy seat side by side is
[a] ${\displaystyle \frac{1}{2}}$
[b] ${\displaystyle \frac{1}{4}}$
[c] ${\displaystyle \frac{1}{3}}$
[d] ${\displaystyle \frac{1}{6}}$

Answer 1

Hint: Probability of event E = ${\displaystyle \frac{n(E)}{n(S)}}={\displaystyle \frac{\text{Favourable cases}}{\text{Total number of cases}}}$ where S is called the sample space of the random experiment. Find n (E) and n (S) and use the above formula to find the probability. Use the grouping method to find the values of n(E) and use the fact that arranging n elements in a row is given by n!.

Complete step-by-step answer:
Let E be the event: The two girls occupy seats side by side.
Let us consider the girls as a single group. Hence we have four boys and one group to be arranged which can be done in 5! ways. But the girls inside the group can internally arrange themselves in 2! ways.
Hence the total number of ways in which four boys and two girls can be seated so that the girls are seated side by side is given by 5!2! ways
Hence, we have n (E) = 5!2! = 240.
The total number of ways in which we can arrange four boys and two girls without any restriction is given by 6! =720.
Hence, we have n (S) =720
Hence, P (E) = ${\displaystyle \frac{240}{720}}={\displaystyle \frac{1}{3}}$
Hence option [c] is correct.

Note: [1] It is important to note that arranging at random is important for the application of the above problem. If the arrangement is not random, then there is a bias factor in arranging, and the above formula is not applicable. In those cases, we use the conditional probability of an event.
[2] The probability of an event always lies between 0 and 1
[3] The sum of probabilities of an event E and its complement E’ = 1
i.e. $P(E)+P(E^{\prime })=1$
Hence, we have $P(E^{\prime })=1-P(E)$. This formula is applied when it is easier to calculate P(E’) instead of P(E).
Here E’: The girls are not together.
We find n(E’) by gap method.
First, we arrange four boys. This can be done in 4! ways.
Now there are five gaps in between 4 boys as shown
X B X B X B X B X
We seat the girls in those gaps. For that, we need to select two places at which the girls have to be placed and the arrangement of the girls in those two places. Selection can be done in ${}^5{C}_2$ ways, and the arrangement can be done in $2!$ ways. Hence the total number of ways of seating the girls so that they are not together $=4{!}^5{C}_{2}2!$
Using ${}^n{C}_r={\displaystyle \frac{n!}{r!(n-r)!}}$, we get
${}^5{C}_2={\displaystyle \frac{5!}{3!2!}}={\displaystyle \frac{120}{12}}=10$.
Hence the total number of ways of seating of four boys and two girls so that the girls are not together = 480.
Hence P(E’) $={\displaystyle \frac{480}{720}}={\displaystyle \frac{2}{3}}$.
Hence we have P(E) $={\displaystyle \frac{1}{3}}$.