Name: Random Variables in Probability Theory
Uploaded: 2019-07-19
Duration: 8 min 57 s
Description: What do we mean by random variables? Watch this video to learn about random variables alongside continuous random variables and discrete random variables wit...

Question

A multiple choice test has

15

questions, and each question has four possible answers. There is only one correct answer per question. If

X

is "the number of questions a student guesses correctly", find: the probability that, if a student answers the test purely by guessing, they will achieve the pass mark of

10

or more.

Accepted Answer

Given, $X$ is " the number of questions a student guesses correctly",

Now, total number of questions are $15$ . i.e., number of trials can be considered to be $15$ .

Also, the trials are independent. And, the number of mutually exclusive events are $2$ i.e., either the answer is correct or wrong. And, the probability of success $= p =$ Probability of getting the correct answer for one question $= \frac{1}{4} = 0.25$ .

Thus, the given experiment is satisfying all the conditions of Binomial distribution.

$\Rightarrow X$ is a binomially distributed random variable with $n = 15$ , and $p = 0.25$ i.e., $X ~ B (15, 0.25)$

We need to find: $P (X \geq 10)$

We know that the probability of getting $r$ success out of $n$ independent trials, when $p$ is the probability of success for each trial is

$P (X = r) = (\begin{array}{l} n \\ r \end{array}) p^{r} (1 - p)^{n - r} = (\begin{array}{l} n \\ r \end{array}) p^{r} (q)^{n - r}$

Now,

$P (X \geq 10) = P (X = 10) + P (X = 11) + P (X = 12) + P (X = 13) + P (X = 14) + (X = 15)$

$= {}^{15}C_{10} {(\frac{1}{4})}^{10} {(\frac{3}{4})}^{5} + {}^{15}C_{11} {(\frac{1}{4})}^{11} {(\frac{3}{4})}^{4} + {}^{15}C_{12} {(\frac{1}{4})}^{12} {(\frac{3}{4})}^{3} + {}^{15}C_{13} {(\frac{1}{4})}^{13} {(\frac{3}{4})}^{2} + {}^{15}C_{14} {(\frac{1}{4})}^{14} (\frac{3}{4}) + {}^{15}C_{15} {(\frac{1}{4})}^{15} (\frac{3}{4})$

$= \frac{15!}{10! \cdot 5!} {(\frac{1}{4})}^{10} {(\frac{3}{4})}^{5} + \frac{15!}{11! \cdot 4!} {(\frac{1}{4})}^{11} {(\frac{3}{4})}^{4} + \frac{15!}{12! \cdot 3!} {(\frac{1}{4})}^{12} {(\frac{3}{4})}^{3} + \frac{15!}{13! \cdot 2!} {(\frac{1}{4})}^{13} {(\frac{3}{4})}^{2} + \frac{15!}{14!} {(\frac{1}{4})}^{14} (\frac{3}{4}) + {(\frac{1}{4})}^{15} {(\frac{3}{4})}^{0}$

$= 3003 {(\frac{1}{4})}^{10} {(\frac{3}{4})}^{5} + 1365 {(\frac{1}{4})}^{11} {(\frac{3}{4})}^{4} + 455 {(\frac{1}{4})}^{12} {(\frac{3}{4})}^{3} + 105 {(\frac{1}{4})}^{13} {(\frac{3}{4})}^{2} + 15 {(\frac{1}{4})}^{14} (\frac{3}{4}) + {(\frac{1}{4})}^{15}$

$= {(\frac{1}{4})}^{10} (\frac{3}{4}) [3003 {(\frac{3}{4})}^{4} + 1365 (\frac{1}{4}) {(\frac{3}{4})}^{3} + 455 {(\frac{1}{4})}^{2} {(\frac{3}{4})}^{2} + 105 {(\frac{1}{4})}^{3} {(\frac{3}{4})}^{1} + 15 {(\frac{1}{4})}^{4} + {(\frac{1}{4})}^{5}]$

$= 0.000795$

Hence, $P (X \geq 10) = 0.000795$

Important Questions on Valid Comparisons and Informed Decisions: Probability Distributions

Learn and Prepare for any exam you want !