# Probability Question

#### cameron80

##### New member
Tim participates in a multiple choice examination consisting of 100 questions. No marks are deducted
for incorrect answers and one mark is awarded for each correct answer. Tim calculates there are
65 questions he is sure of getting right. Of the remaining 35, there is a one in three (1/3) chance that
any given question is answered correctly.

What is the probability, to three decimal places, of Tim passing the exam if the pass mark is
(a) 70
(b) 75
(c) 80?

#### soroban

##### Elite Member
Hello, cameron80!

I'll do part (a) . . .

Tim participates in a multiple choice examination consisting of 100 questions.
No marks are deducted for incorrect answers and one mark is awarded for each correct answer.
Tim calculates there are 65 questions he is sure of getting right.
Of the remaining 35, there is a 1/3 probabiity that any given question is answered correctly.

What is the probability, to three decimal places, of Tim passing the exam if the pass mark is:

. . $$\displaystyle (a)\;70$$

To get a grade of at least 70, he needs at least 5 more correct answers
. . from among the other 35 questions.

He will fail if he gets 4 or less correct.
We will find the probability that he will fail.
This will happen if he gets 0, 1, 2, 3, or 4 correct answers.

$$\displaystyle \begin{array}{ccccc}\text{0 right:} & {35\choose0}(\frac{1}{3})^0(\frac{2}{3})^{35} &=& 1\cdot\frac{2^{35}}{3^{35}} \\ \text{1 right:} & {35\choose1}(\frac{1}{3})^1(\frac{2}{3})^{34} &=& 35\cdot\frac{2^{34}}{3^{35}} \\ \text{2 right:} & {35\choose2}(\frac{1}{3})^2(\frac{2}{3})^{33} &=& 595\cdot\frac{2^{32}}{3^{35}} \\ \text{3 right:} & {35\choose3}(\frac{1}{3})^3(\frac{2}{3})^{32} &=& 6,\!545\cdot\frac{2^{32}}{3^{35}} \\ \text{4 right:} & {35\choose4}(\frac{1}{3})^4(\frac{2}{3})^{31} &=& 52,\!360\cdot\frac{2^{31}}{3^{35}} \end{array}$$

$$\displaystyle P(\text{4 or less}) \;=\;\dfrac{2^{35}}{3^{35}} + 35\dfrac{2^{34}}{3^{35}} + 595\dfrac{2^{33}}{3^{35}} + 6,\!545\dfrac{2^{32}}{3^{35}} + 52,\!360\dfrac{2^{31}}{3^{35}}$$

. . . . . . . . . .$$\displaystyle =\;\dfrac{2^{31}}{3^{35}}\left[2^4 + 35\cdot2^3 + 595\cdot 2^2 + 6,\!545\cdot 2 + 52,\!360\right]$$

. . . . . . . . . .$$\displaystyle =\;\dfrac{2^{31}}{3^{35}}(68,\!126) \;=\;\left(\dfrac{2}{3}\right)^{31}\dfrac{68,\!126}{3^4} \;=\;0.002924145$$

$$\displaystyle P(\text{4 or less}) \;\approx\;0.003$$

$$\displaystyle \text{Therefore: }\(\text{5 or more}) \;=\;P(\text{pass}) \;=\;1 - 0.003 \;=\;0.997$$