Question

For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve the problem is $\dfrac{4}{5}$, then the probability that he is unable to solve less than two problems is

& \text{(A) }\dfrac{316}{25}{{\left( \dfrac{4}{5} \right)}^{48}} \\\ & (\text{B) }\dfrac{54}{5}{{\left( \dfrac{4}{5} \right)}^{48}} \\\ & (\text{C) }\dfrac{164}{25}{{\left( \dfrac{1}{5} \right)}^{48}} \\\ & (\text{D) }\dfrac{201}{5}{{\left( \dfrac{1}{5} \right)}^{49}} \\\ \end{aligned}$$

Answer 1

Hint: We should know that the probability of getting “k” success in “n” independent trials is equal to $P(k)$ where $P(k){{=}^{n}}{{C}_{k}}{{p}^{k}}{{q}^{n-k}}$, p is the probability of getting success in an independent event and q is the probability of not getting success in an independent event. Let us assume that the probability that a candidate should solve less than 2 problems is $P(X<2)$, the probability that a candidate can solve only one problem is $P(X=1)$and the probability that a candidate cannot any problem is $P(X=0)$. We know that the probability to solve less than two problems is equal to the sum of probability to solve one problem and probability that he cannot solve even one problem. Let us assume this equation as equation (1). We know that p is probability that the problem cannot be solved and q is probability that the problem can be solved. Assume the value of p as equation (2) and the value of q as equation (3). By using equation (1), equation (20 and equation (3), we can find the value of $P(X<2)$. This gives us the probability of making less than two solutions as incorrect/

Complete step-by-step answer:
In the question, it was given that the candidate has given fifty problems to solve.
We have given that the probability that the candidate can solve the problem is $\dfrac{4}{5}$.
From the question, it is given that we have to find the probability that the candidate should not solve less than 2 problems.
We know that the probability of getting “k” success in “n” independent trials is equal to $P(k)$ where $P(k){{=}^{n}}{{C}_{k}}{{p}^{k}}{{q}^{n-k}}$, p is the probability of getting success in an independent event and q is the probability of not getting success in an independent event.
Let us assume that the probability that a candidate should solve less than 2 problems is $P(X<2)$, the probability that a candidate can solve only one problem is $P(X=1)$and the probability that a candidate cannot any problem is $P(X=0)$.
We know that the probability to solve less than two problems is equal to the sum of probability to solve one problem and probability that he cannot solve even one problem.
So, we can get
$\Rightarrow P(X<2)=P(X=0)+P(X=1).....(1)$
We know that p is probability that the problem cannot be solved and q is probability that the problem can be solved.
So, we can get

& p=\dfrac{1}{5}....(2) \\\ & q=1-p \\\ & \Rightarrow q=1-\dfrac{1}{5} \\\ & \Rightarrow q=\dfrac{4}{5}....(3) \\\ \end{aligned}$$ From equation (1), equation (2) and equation (3) we get $$\begin{aligned} & \Rightarrow P(X<2)=P(X=0)+P(X=1) \\\ & \Rightarrow P(X<2){{=}^{50}}{{C}_{0}}{{\left( \dfrac{1}{5} \right)}^{0}}{{\left( \dfrac{4}{5} \right)}^{50}}{{+}^{50}}{{C}_{1}}{{\left( \dfrac{1}{5} \right)}^{1}}{{\left( \dfrac{4}{5} \right)}^{49}} \\\ & \Rightarrow P(X<2)={{\left( \dfrac{4}{5} \right)}^{50}}+50\left( \dfrac{1}{5} \right){{\left( \dfrac{4}{5} \right)}^{49}} \\\ & \Rightarrow P(X<2)=\left( \dfrac{54}{5} \right){{\left( \dfrac{4}{5} \right)}^{48}} \\\ \end{aligned}$$ So, we can conclude that the probability that he is unable to solve less than two problems is equal to $$\left( \dfrac{54}{5} \right){{\left( \dfrac{4}{5} \right)}^{48}}$$. Hence, option (B) is correct. Note: Students need to read the question carefully. Because if this question can be read that we have to find the probability to solve two problems. If this mistake is followed, the student will assume the event X as that the candidate solved the problem correctly. $$\Rightarrow P(X<2)=P(X=0)+P(X=1).....(1)$$ So, we can get $$\begin{aligned} & p=\dfrac{4}{5}....(2) \\\ & q=1-p \\\ & \Rightarrow q=1-\dfrac{4}{5} \\\ & \Rightarrow q=\dfrac{1}{5}....(3) \\\ \end{aligned}$$ From equation (1), equation (2) and equation (3) we get $$\begin{aligned} & \Rightarrow P(X<2)=P(X=0)+P(X=1) \\\ & \Rightarrow P(X<2){{=}^{50}}{{C}_{0}}{{\left( \dfrac{4}{5} \right)}^{0}}{{\left( \dfrac{1}{5} \right)}^{50}}{{+}^{50}}{{C}_{1}}{{\left( \dfrac{4}{5} \right)}^{1}}{{\left( \dfrac{1}{5} \right)}^{49}} \\\ & \Rightarrow P(X<2)={{\left( \dfrac{1}{5} \right)}^{50}}+50\left( \dfrac{4}{5} \right){{\left( \dfrac{1}{5} \right)}^{49}} \\\ & \Rightarrow P(X<2)=(21){{\left( \dfrac{1}{5} \right)}^{50}} \\\ \end{aligned}$$ This misconception gives us a wrong answer. So, students should read the question properly.