Question

If a test has three multiple choice questions that offer five possible answers, determine the probability that a student will get one out of three questions correct by guessing on all of them.
(a) $\dfrac{49}{125}$
(b) $\dfrac{48}{125}$
(c) $\dfrac{48}{145}$
(d) $\dfrac{49}{145}$

Answer 1

Hint: We will first find total combination of multiple answer which will be $5\times 5\times 5=125$ . then we will find probability of correct answer $P\left( c \right)=\dfrac{1}{5}$ and subtracting this from 1, we will get probability of incorrect answer i.e. $P\left( I \right)=\dfrac{4}{5}$ . Now, we will select 1 question out of 3 questions which is correct i.e. ${}^{3}{{C}_{1}}$ . Along with this we will multiply the probability of getting the correct answer one time and that of incorrect answer 2 times and on solving, we will get the answer. Formula we will be using is ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ .

Complete step-by-step answer:
We are given that there are 3 multiple choice questions and each question has 5 options for answer. So, the total possible multiple combination of selecting the answer for all 3 questions will be $5\times 5\times 5=125$ .
Now, we know that out of 5 options only one is correct. So, we can say that the probability of the correct answer is $\dfrac{1}{5}$ . So, that incorrect answer will be $1-\dfrac{1}{5}=\dfrac{4}{5}$ because we know that maximum probability is 1. So, we have subtracted probability of correct answer from 1 in order to get probability of incorrect answer.
Probability of correct answer $P\left( c \right)=\dfrac{1}{5}$
Probability of incorrect answer $P\left( I \right)=\dfrac{4}{5}$
Now, we have to find the probability of one correct question out of 3. So, we can write it as ${}^{3}{{C}_{1}}$ . Also, that one question is correct so, probability of getting the correct answer is to be multiplied and for the remaining 2 questions which are incorrect also their incorrect answer probability should be multiplied. So, we get as
$={}^{3}{{C}_{1}}\times P\left( c \right)\times P\left( I \right)$
On putting values and using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ , we get as
$=\dfrac{3!}{\left( 3-1 \right)!1!}\times \dfrac{1}{5}\times \dfrac{4}{5}\times \dfrac{4}{5}$
(we have multiplied $P\left( I \right)=\dfrac{4}{5}$ 2 times because we have 2 incorrect questions)
On further solving, we get as
$=\dfrac{3\times 2!}{2!1!}\times \dfrac{16}{125}$
On simplification we get as,
$=\dfrac{3\times 16}{125}=\dfrac{48}{125}$
Thus, option (b) is the correct answer.

Note: Remember that we have to select only one correct answer, so we have 2 incorrect questions with us. So, we have to multiple probability of correct answer one time and that of incorrect answer twice. Students make mistake here and multiply both probability one time only and get answer wrong i.e. $=\dfrac{3!}{\left( 3-1 \right)!1!}\times \dfrac{1}{5}\times \dfrac{4}{5}$ . on solving this, we will get incorrect answer. So, do not make this mistake.