Question

In a multiple choice question, there are four alternative answers, of which one or more are correct. A candidate decides to tick the answers at random. If he is allowed u to 3 choices to answer the question, the probability that he will get arks in the question is
(a)$\dfrac{1}{15}$
(b)$\dfrac{621}{3375}$
(c)$\dfrac{1}{5}$
(d)$\dfrac{4}{15}$

Answer 1

Hint: To solve this question, first we will find the number of ways to answer a question by given data in question. then we will find all the situations for getting marks in question as he is allowed up to 3 choices then using properties of combination and factorial function we will solve each case and hence, on adding we will get the answer.

Complete step-by-step answer:
Now, as each question of multiple choice can have one or more correct answers, so may be out of 4 options only 1 option is correct or out of 4 options only 2 option is correct or out of 4 options only 3 option is correct or all four options are correct.
We know that selecting r objects from total n objects is case of combination and denotes as $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
So, The total number of ways to answer the question will be,
 $^{4}{{C}_{1}}{{+}^{4}}{{C}_{2}}{{+}^{4}}{{C}_{3}}{{+}^{4}}{{C}_{4}}$
We know that, $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$,
So, on expanding, we get
$=\dfrac{4!}{1!\left( 4-1 \right)!}+\dfrac{4!}{2!\left( 4-2 \right)!}+\dfrac{4!}{3!\left( 4-3 \right)!}+\dfrac{4!}{4!\left( 4-4 \right)!}$, where x ! is a factorial function which is evaluated as $x!=x(x-1)(x-2)(x-3)....3.2.1$ where $x\ge 0$ and also, we know that $^{n}{{C}_{n}}=1$, where $n\ge 0$ .
On simplifying, we get
$=\dfrac{4\times 3!}{1!\left( 3 \right)!}+\dfrac{4\times 3\times 2!}{2!\left( 2 \right)!}+\dfrac{4\times 3!}{3!\left( 1 \right)!}+\dfrac{4!}{4!\left( 0 \right)!}$, we know that 0! = 1
So, on solving we get
= 4 + 6 + 4 + 1
So, The total number of ways to answer the question will be 15.
The, probability of getting question correct $=\dfrac{1}{15}$
Now, the candidate may be correct on the first or second or third chance as in question it is given that he is allowed u to 3 choices to answer the question, so these probabilities are mutually exclusive as mutually exclusive events are those events which cannot happen simultaneously.
So, P(getting marks) = P( correct answer in first attempt ) + P ( correct answer in second attempt ) + P(correct answer in third attempt )……( i )
So, P( correct answer in first attempt ) $=\dfrac{1}{15}$
P ( correct answer in second attempt ) $=\dfrac{14}{15}\times \dfrac{1}{14}$, as in first attempt getting answer wrong means probability becomes $1-\dfrac{1}{15}=\dfrac{14}{15}$ and as first attempt is already occurred, so probability that question gets correct in second attempt will be $\dfrac{1}{14}$
P ( correct answer in third attempt ) \[=\dfrac{14}{15}\times \dfrac{13}{14}\times \dfrac{1}{13}\], as in first attempt getting answer wrong means probability becomes $1-\dfrac{1}{15}=\dfrac{14}{15}$ and in second attempt answer gets wrong so probability becomes $1-\dfrac{1}{14}=\dfrac{13}{14}$ and as two attempts are already done, probability so that question gets correct in third attempt will be $\dfrac{1}{13}$
Putting values of P( correct answer in first attempt ), P ( correct answer in second attempt ) and P(correct answer in third attempt ) in equation ( i ), we get
P(getting marks) $=\dfrac{1}{15}+\dfrac{14}{15}\times \dfrac{1}{14}+\dfrac{14}{15}\times \dfrac{13}{14}\times \dfrac{1}{13}$
On simplifying, we get
$=\dfrac{1}{15}+\dfrac{1}{15}+\dfrac{1}{15}$
On solving, we get
$\begin{align}
  & =\dfrac{3}{15} \\
 & =\dfrac{1}{5} \\
\end{align}$

So, the correct answer is “Option c”.

Note: Always remember that if probability of happening of an event is x then, probability of non happening of an event will be 1 – x. Always remember that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $x!=x(x-1)(x-2)(x-3)....3.2.1$. Try to avoid calculation mistakes.