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# In a multiple choice test, there are 6 questions. Four alternative answers are given for each question, of which only one answer is correct. If a student answers all the questions by choosing one answer for each question, then the number of ways of getting exactly 4 correct answer isA) ${4^6} - {4^2}$ B) $135$ C) $9$ D) $120$

Last updated date: 13th Jun 2024
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Hint:
Here, we will use the concept of the binomial distribution to find the number of ways. The binomial distribution is a distribution that is used to find the number of times success occurred in a trial.
Formula Used:
We will use the following formulas:
1) Binomial distribution is given by the formula $P\left( {x = r} \right) = {}^n{C_r}{\left( p \right)^r}{\left( q \right)^{n - r}}$, where $p$ is the success, $q$ is the failure, ${}^n{C_r}$ is the number of combinations, $n$ is the number of trials and $r$ is the number of required success.
2) Number of combinations is given by the formula ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.

Complete step by step solution:
We are given 6 questions in a multiple choice test.
Total number of questions $= 6$.
Number of Questions answered correctly $= 4$.
Out of the four alternative answers given for each question, only one answer is correct.
We will find the number of ways of getting exactly 4 correct answer by using the binomial distribution.
We are given that Number of trials $n = 6$, Number of required success $r = 4$, Number of correct answers in a question $p = 1$, Number of wrong answers in a question $q = 3$
Now, substituting $n = 6$, $r = 4$, $p = 1$ and $q = 3$ in the formula $P\left( {x = r} \right) = {}^n{C_r}{\left( p \right)^r}{\left( q \right)^{n - r}}$, we get
$\Rightarrow P\left( {x = 4} \right) = {}^6{C_4}{\left( 1 \right)^4}{\left( 3 \right)^{6 - 4}}$
$\Rightarrow P\left( {x = 4} \right) = {}^6{C_4}{\left( 1 \right)^4}{\left( 3 \right)^2}$
Now using the formula ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get
$\Rightarrow P\left( {x = 4} \right) = \dfrac{{6!}}{{4!\left( {6 - 4} \right)!}}{\left( 1 \right)^4}{\left( 3 \right)^2}$
Computing the factorial, we get
$\Rightarrow P\left( {x = 4} \right) = \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1 \times 2 \times 1}}\left( 1 \right)\left( 9 \right)$
Simplifying the expression, we get
$\Rightarrow P\left( {x = 4} \right) = \left( {15} \right) \times \left( 9 \right)$
Multiplying the terms, we get
$\Rightarrow P\left( {x = 4} \right) = 135$
Therefore, the number of ways of getting exactly 4 correct answers is 135.

Thus, Option (B) is the correct answer.

Note:
We know that the number of ways can be calculated by using permutations and combinations. Permutation is a way or method of arranging a set of elements and the order of arrangement matters. Combination is a method of selecting an element from a given set of elements but order does not matter. The binomial distribution is applicable only if the event has only two favorable outcomes that are either success or failure. This method can also be used when the given events are in probability.