Question

On a multiple-choice examination with three possible answers for each of five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

Answer 1

For solving this problem we use Bernoulli’s trails. Here, we know that each question has two rates either success or failure in which we are given that four or more answers are correct. This is called Bernoulli’s trail. We use formula if ‘p’ and ‘q’ are probabilities of success or failure respectively and ‘n’ is total number of questions we can take probability as $P\left( X=k \right)={}^{n}{{C}_{k}}{{p}^{k}}{{q}^{n-k}}$. Here, we need to find the probability of getting four or more correct answers as
$P\left( X\ge 4 \right)=P\left( X=4 \right)+P\left( X=5 \right)$.

Complete step-by-step solution
Let us assume that $X$ be the event of getting more than four correct answers.
We know that the probability of getting a correct answer is $p=\dfrac{1}{3}$ because there will be only one correct answer out of three options.
Now, let us find the probability of getting wrong answer as

& \Rightarrow q=1-p \\\ & \Rightarrow q=1-\dfrac{1}{3}=\dfrac{2}{3} \\\ \end{aligned}$$ Since the given event is Bernoulli’s trail we use the formula $$P\left( X=k \right)={}^{n}{{C}_{k}}{{p}^{k}}{{q}^{n-k}}$$ to find the probability of getting four or more answers as correct. Here, we let us find probability of getting four or more correct answers as $$\begin{aligned} & \Rightarrow P\left( X\ge 4 \right)=P\left( X=4 \right)+P\left( X=5 \right) \\\ & \Rightarrow P\left( X\ge 4 \right)={}^{5}{{C}_{4}}{{\left( \dfrac{1}{3} \right)}^{4}}{{\left( \dfrac{2}{3} \right)}^{5-4}}+{}^{5}{{C}_{5}}{{\left( \dfrac{1}{3} \right)}^{5}}{{\left( \dfrac{2}{3} \right)}^{5-5}} \\\ \end{aligned}$$ We know that $${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$$. By using this formula in above equation we will get $$\begin{aligned} & \Rightarrow P\left( X\ge 4 \right)=\left( 5 \right)\left( \dfrac{2}{243} \right)+\left( 1 \right)\left( \dfrac{1}{243} \right) \\\ & \Rightarrow P\left( X\ge 4 \right)=\dfrac{11}{243} \\\ \end{aligned}$$ **Therefore the probability of getting four or more correct answers just by guessing is $$\dfrac{11}{243}$$.** **Note:** Students may make mistakes in applying the Bernoulli’s trial formula. That is instead of taking the formula as $$P\left( X=k \right)={}^{n}{{C}_{k}}{{p}^{k}}{{q}^{n-k}}$$ they may take $$P\left( X=k \right)={}^{n}{{C}_{k}}{{q}^{k}}{{p}^{n-k}}$$ which results in wrong answer. Here, ‘k’ should be the power of ‘p’ not ‘q’. Also, we are asked to find the probability of getting four or more correct answers. We need to add both $$P\left( X=4 \right),P\left( X=5 \right)$$.