Chapter 10.1, Problem 40E

Solution

(a).

To calculate: The probability that the student will be able to answer all $8$ questions if a given student knows how to answer $14$ of the questions.

The probability that the student will be able to answer all $8$ is $30.7\%$ .

Given information:

Professor Indiana Jones gives his class a list of $20$ study questions, from which he will select $8$ to be answered on the final exam.

Calculation:

The chance that a student that knows  $14$  out of  $20$  questions will know all  $8$  questions on the test is:

  $P=\frac{C{}_4}{C{}_8}$

Now, apply the combination formula to calculate $\frac{C{}_4}{C{}_8}$ .

  ${}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}$ where, $n$ is the total number of object and $r$ is the number of choosing object from the set.

Substitute the values into the formula:

  $\begin{array}{l}P=\frac{\frac{20!}{4!\left(20-4\right)!}}{\frac{20!}{8!\left(20-8\right)!}}\\ =\frac{\frac{20\times 19\times 18\times 17\times 16!}{4\times 3\times 2\times 1\times 16!}}{\frac{20\times 19\times 18\times 17\times 16\times 15\times 14\times 13\times 12!}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1\times 12!}}\\ =\frac{\frac{20\times 19\times 18\times 17}{4\times 3\times 2}}{\frac{20\times 19\times 18\times 17\times 16\times 15\times 14\times 13}{8\times 7\times 6\times 5\times 4\times 3\times 2}}\\ =\frac{38760}{125970}\\ \approx 0.307\end{array}$

Hence, if a given student knows how to answer $14$ of the questions, then the probability that the student will be able to answer all $8$ is $30.7\%$ .

(b).

To calculate: The probability that the student will be able to answer exactly $5$ questions if a given student knows how to answer $14$ of the questions.

The probability that the student will be able to answer exactly $5$ questions is $31.8\%$ .

Given information:

Professor Indiana Jones gives his class a list of $20$ study questions, from which he will select $8$ to be answered on the final exam.

Calculation:

The chance that a student that know  $14$  out of  $20$  questions will know  $5$  questions on the test is the product of the number of combinations of  $5$  questions that the student knows and  $3$ number of combinations of  $3$  he does not know divided by the number of possible combinations of questions:

  $P=\frac{C{}_5\cdot C{}_3}{C{}_8}$

Now, apply the combination formula to calculate $\frac{C{}_5\cdot C{}_3}{C{}_8}$ .

  ${}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}$ where, $n$ is the total number of object and $r$ is the number of choosing object from the set.

Substitute the values into the formula:

  $\begin{array}{l}P=\frac{\frac{14!}{5!\left(14-5\right)!}\times \frac{6!}{3!\left(6-3\right)!}}{\frac{20!}{8!\left(20-8\right)!}}\\ =\frac{\frac{14\times 13\times 12\times 11\times 10\times 9!}{5\times 4\times 3\times 2\times 1\times 9!}\times \frac{6\times 5\times 4\times 3!}{3\times 2\times 1\times 3!}}{\frac{20\times 19\times 18\times 17\times 16\times 15\times 14\times 13\times 12!}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1\times 12!}}\\ =\frac{2002\times 20}{125970}\\ \approx 0.318\end{array}$

Hence, if a given student knows how to answer $14$ of the questions, then the student will be able to answer exactly $5$ questions is $31.8\%$

(c).

To calculate: The probability that the student will be able to answer at least $6$ questions if a given student knows how to answer $14$ of the questions.

The probability the student will be able to answer at least $6$ questions is $\frac{176}{323}$ .

Given information:

Professor Indiana Jones gives his class a list of $20$ study questions, from which he will select $8$ to be answered on the final exam.

Calculation:

The number of ways the professor can select $8$ questions out of $20$ questions is $C{}_8$

?, which will the denominator of the desired probability.

The numerator of the desired probability is the number of ways the student answers at least 6 questions correctly. There are three cases: a) exactly 6 questions correctly, b) exactly 7 questions correctly and c) all 8 questions correctly.

The number of ways the student answers exactly 6 questions correctly is $C{}_6\times C{}_2$ since if she answers 6 questions correctly out of the 14 questions she knows, she must answer 2 questions incorrectly out the 6 questions she doesn't know. Similarly, the number of ways the student answers exactly 7 and all 8 questions correctly is $C{}_7\times C{}_1$ and $C{}_8\times C{}_0$ , respectively. Therefore, the desired probability is:

  $P=\frac{C{}_6\times C{}_2\times C{}_7\times C{}_1\times C{}_8\times C{}_0}{C{}_8}$

Now, apply the combination formula to calculate $\frac{C{}_5\cdot C{}_3}{C{}_8}$ .

  ${}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}$ where, $n$ is the total number of object and $r$ is the number of choosing object from the set.

Which can be simply to $\frac{176}{323}$ or approximately to $\mathrm{0.545.}$

Hence, questions if a given student knows how to answer $14$ of the question then probability the student will be able to answer at least $6$ questions is $\frac{176}{323}$ .