Question

A statistics textbook chapter contains $63$ exercises, $9$ of which are essay questions: student is assigned $13$ problems

(a) What is the probability that none of the questions are essay? (Round your answer to decimal places )

(b) What is the probability that at least one is essay? (Round your answer to $4$ decimal places) Probability

(c) What is the probability that two or more are essay? (Round your answer to $4$ decimal places )

Answer 1

(a) . Probability that none of the questions are essay.

It is given that $9$ questions are of essay.

Total exercise are $63$.

If none of the the assign problem are essay then remaining problems are $63-9=54$.

Step -1: Find the probability that none of the questions are essay:

$\mathrm{P}(\mathrm{None}\mathrm{of}\mathrm{the}\mathrm{question}\mathrm{are}\mathrm{essay})=\left(\frac{{}^{9}C_0\times {}^{54}C_{13}}{{}^{63}C_{13}}\right)$.

${}^{54}C_{13}$ denotes the number of ways to chose $13$ question out of remaining $54$

${}^{9}C_0$denotes the $9$ questions apart from $54$ questions.

${}^{63}C_{13}$denotes the total number of ways assigning $13$ problem out of $63$ .

Step 2.Simplifying the expression:

Apply the combination formula ${}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{r})!\times \mathrm{r}!}$.

$$\begin{gathered} \mathrm{P}(\mathrm{None}\mathrm{of}\mathrm{the}\mathrm{question}\mathrm{is}\mathrm{essay})=\left(\frac{{\displaystyle \frac{9!}{(9-0)!\times 0!}\times {}^{54}C_{13}}}{{}^{63}C_{13}}\right) \\ =\left(\frac{{\displaystyle \frac{9!}{9!}\times \frac{54!}{(54-13)!\times 13!}}}{{}^{63}C_{13}}\right) \\ =\left(\frac{{\displaystyle \frac{54!}{41!\times 13!}}}{{\displaystyle \frac{63!}{(63-13)!\times 13!}}}\right) \\ =\left(\frac{54!}{41!\times 13!}\times \frac{50!\times 13!}{63!}\right) \\ =0.10585 \end{gathered}$$

Hence the probability that none of the questions are essay is $0.10585$.

(b) Find the probability of at least one question is essay:

$$\begin{gathered} \mathrm{P}(\mathrm{At}\mathrm{least}\mathrm{one}\mathrm{question}\mathrm{is}\mathrm{of}\mathrm{essay})=1-\mathrm{P}(\mathrm{None}\mathrm{of}\mathrm{the}\mathrm{question}\mathrm{is}\mathrm{essay} \\ =1-0.10585 \\ =0.8945 \end{gathered}$$

(c ) Find the probability that two or more are essay:

$\mathrm{P}(\mathrm{Two}\mathrm{or}\mathrm{more}\mathrm{are}\mathrm{essay})=1-\mathrm{P}(\mathrm{None}\mathrm{is}\mathrm{essay})-\mathrm{P}(\mathrm{Exactly}\mathrm{one}\mathrm{is}\mathrm{essay})$.

In order to find the probability that two or more are essay it needs to find the probability of assigning exactly one essay problem.

Step- 1: Find the probability of assigning exactly one essay problem:

$\mathrm{P}(\mathrm{Exactly}\mathrm{one}\mathrm{essay}\mathrm{problem})=\frac{{}^{9}C_1\times {}^{54}C_{12}}{{}^{63}C_{12}}$. .

${}^{9}C_1$denotes the number of ways that exactly one problem can be assigned out of $9$ problem.

${}^{54}C_{12}$ denotes the number of ways the remaining $12$ problems can be assigned out of $54$ problem.

${}^{63}C_{13}$denotes the same as explained above.

$$\begin{gathered} \mathrm{P}(\mathrm{Exactly}\mathrm{one}\mathrm{essay}\mathrm{problem})=\left(\frac{{\displaystyle \frac{9!}{(9-1)!\times 1!}}\times {\displaystyle \frac{54!}{(54-12)!\times 12!}}}{{}^{63}C_{12}}\right) \\ =\left(\frac{{\displaystyle \frac{9!}{8!}}\times {\displaystyle \frac{54!}{42!\times 12!}}}{{}^{63}C_{12}}\right) \\ =0.2948 \end{gathered}$$

Step -2: Find the probability that two or more are essay:

$\mathrm{P}(\mathrm{Two}\mathrm{or}\mathrm{more}\mathrm{are}\mathrm{essay})=1-\mathrm{P}(\mathrm{None}\mathrm{is}\mathrm{essay})-\mathrm{P}(\mathrm{Exactly}\mathrm{one}\mathrm{is}\mathrm{essay})$.

Substitute the known values.

$$\begin{gathered} \mathrm{P}(\mathrm{Two}\mathrm{or}\mathrm{more}\mathrm{are}\mathrm{essay})=1-0.1058-0.2948 \\ =1-0.4006 \\ =0.5994 \end{gathered}$$

Probability that two or more are essay is $0.5994$.

The probability that none of the questions are essay is $0.1058$,the probability that at least one is essay is $0.2948$ and the probability that two or more are essay is $0.5994$.