Chapter 12, Problem 1PTTS

To determine

To find:the different groups of 5 students that teacher choose include me and included me and friend and the probability that me and my friend will be in the same group.

Answer to Problem 1PTTS

1. Thenumber of groups included me is $1001$ .
2. The number of different groups included me and my friend is $715$ .
3. The probability to me and my friend in the same group is $\frac{858}{1001}$ .

Explanation of Solution

Given :

Total students = $15$

Calculation:

1. To find the different groups included me, I am selected so we have to choose 4 students from $15-1$so we use the combination,
Groups = $C{}_4$

  = $\begin{array}{l}\frac{14!}{4!·\left(14-4\right)!}\text{ [substitute n by 14 and r by 4]}\\ \frac{14!}{4!·10!}\text{ [subtract]}\\ \frac{14·13·12·11·10!}{4·3·2·1·10!}\text{ [expand factorial and divide out common factors]}\\ 1001\text{ [multiply]}\end{array}$

2. When me and my friend selected then we have to choose 4 students from 13 students.
Groups = $C{}_4$

  = $\begin{array}{l}\frac{13!}{4!·\left(13-4\right)!}\text{ [substitute n by 13 and r by 4]}\\ \frac{13!}{4!·9!}\text{ [subtract]}\\ \frac{13·12·11·10·9!}{4·3·2·1·9!}\text{ [expand factorial and divide out common factors]}\\ 715\text{ [multiply]}\end{array}$

3. When me and my friend in the same group we have to choose 3 students from 13 students.

  = $C{}_3$

  = $\begin{array}{l}\frac{13!}{3!·\left(13-3\right)!}\text{ [substitute n by 13 and r by 3]}\\ \frac{13!}{4!·9!}\text{ [subtract]}\\ \frac{13·12·11·10!}{3·2·1·10!}\text{ [expand factorial and divide out common factors]}\\ 858\text{ [multiply]}\end{array}$

Probability = $\frac{858}{1001}$

Hence,

1. The number of groups included me is $1001$ .
2. The number of different groups included me and my friend is $715$ .
3. The probability to me and my friend in the same group is $\frac{858}{1001}$ .