Q. 1.4 - A First Course in Probability

Question

A student is to answer $7$ out of $10$ questions in an examination. How many choices has she? How many if she must answer at least $3$ of the first $5$ questions?

Step-by-Step Solution

Answer

$7$ questions out of $10$ can be selected in $120$ ways and if she must answer at least $3$out of first $5$questions, then $7$ out of $10$ questions can be selected in $110$ ways.

Step 1. Given information.

It is given that,

Total no. of questions $=10$

No. of questions to be answered $=7$

Step 2. Find the choices of selecting 7 out of 10 questions.

A student can select $7$ out of $10$ questions in $\left(\begin{array}{c}10\\ 7\end{array}\right)\text{ways}$

$=\frac{10!}{7!3!}=\frac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}=5\times 3\times 8=120$

Therefore, the no. of choices for selecting $7$ out of $10$ questions is $120$.

Step 3. Find the choices of selecting 7 out of 10 questions, if at least 3 questions are to be selected from first 5 questions.

Let us divide the total questions in two groups, each group containing $5$ questions.

If a student selects $3$ questions from first group then he will select $4$ questions from the second group. So, the no. of choices will be

$$\begin{gathered} \left(\begin{array}{c}5\\ 3\end{array}\right)\left(\begin{array}{c}5\\ 4\end{array}\right)=\frac{5!}{3!2!}\times \frac{5!}{4!1!} \\ =10\times 5 \\ =50 \end{gathered}$$

If a student selects $4$ questions from first group then he will select $3$ questions from the second group. So, the no. of choices will be

$$\begin{gathered} \left(\begin{array}{c}5\\ 4\end{array}\right)\left(\begin{array}{c}5\\ 3\end{array}\right)=\frac{5!}{4!1!}\times \frac{5!}{3!2!} \\ =5\times 10 \\ =50 \end{gathered}$$

If a student selects $5$ questions from first group then he will select $2$ questions from the second group. So, the no. of choices will be

$$\begin{gathered} \left(\begin{array}{c}5\\ 5\end{array}\right)\left(\begin{array}{c}5\\ 2\end{array}\right)=\frac{5!}{5!}\times \frac{5!}{3!2!} \\ =1\times 10 \\ =10 \end{gathered}$$

Therefore, the no. of choices of selecting $7$ out of $10$ questions, if at least $3$ questions are to be selected from first $5$ questions $=50+50+10=110$