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Question: In a multiple choice question there are four alternative answer, of which one or more are correct. A...

In a multiple choice question there are four alternative answer, of which one or more are correct. A candidate will get marks in the question only if he ticks all the correct answers. The candidate decides to tick answer the question, the probability that he will get marks in the question is
A) 12\dfrac{1}{2}
B) 115\dfrac{1}{15}
C) 13\dfrac{1}{3}
D) None of these

Explanation

Solution

As we need to find the probability that the student gets marks, we need to split the probability to how many ways he could have answered and then add it together. We will take the product of all the event probabilities if all of them happen in a single trial and take the sum of probabilities if they happen in different trials. Other probability formula which is needed is,
Probability of failure = 1- probability of success

Complete step-by-step answer:
Step 1: Total ways of answering done by the student from the four options provided is
= Choosing 1 option from 4+ choosing 2 from 4+ choosing 3 from 4+ choosing 4 from 4
=4C1+4C2+4C3+4C4{4_{{C_1}}} + {4_{{C_2}}} + {4_{{C_3}}} + {4_{{C_4}}}
=4+4×31×2+4×3×21×2×3+1 =4+6+4+1 =15  = 4 + \dfrac{{4 \times 3}}{{1 \times 2}} + \dfrac{{4 \times 3 \times 2}}{{1 \times 2 \times 3}} + 1 \\\ = 4 + 6 + 4 + 1 \\\ = 15 \\\
Step 2: Finding the probability of student getting marks by assuming the number of attempts=1,
Probability(student getting marks)=P(getting marks in chance 1)
=115= \dfrac{1}{{15}}
Probability of getting marks for the student =115\dfrac{1}{{15}}

Option B is the correct answer.

Note: *Combination which is used when r elements are been arranged or selected from n objects where order of arrangements doesn’t matter. Denoted by, nCr{n_{{C_r}}}=n!r!(n−r)!\dfrac{{n!}}{{r!\left( {n - r} \right)!}}
*Similar question can be attempted by equating the probability of an option among 4 options to be correct =14\dfrac{1}{4} .
*n!n! denotes the factorial function which is a recursive function defined for positive integers given by the product of all integers less than or equal to n. Implies n!=n×(n−1)!n! = n \times (n - 1)!
*If the number of attempts= k, then the corresponding probability for the student to get marks = k15\dfrac{k}{{15}}