Probability 5 Question 1
1. For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is $\frac{4}{5}$, then the probability that he is unable to solve less than two problem is
(2019 Main, 12 April II)
(a) $\frac{201}{5} \frac{1}{5}$
(b) $\frac{316}{25} \frac{4}{5}$
(c) $\frac{54}{5} \frac{4}{5}$
(d) $\frac{164}{25} \frac{1}{5}{ }^{48}$
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Answer:
Correct Answer: 1. (c)
Solution:
- Given that, there are 50 problems to solve in an admission test and probability that the candidate can solve any problem is $\frac{4}{5}=q$ (say). So, probability that the candidate cannot solve a problem is $p=1-q=1-\frac{4}{5}=\frac{1}{5}$.
Now, let $X$ be a random variable which denotes the number of problems that the candidate is unable to solve. Then, $X$ follows binomial distribution with parameters $n=50$ and $p=\frac{1}{5}$.
Now, according to binomial probability distribution concept
$$ P(X=r)={ }^{50} C _r \frac{1}{5}^{r} \frac{4}{5}^{50-r}, r=0,1, \ldots, 50 $$
$\therefore$ Required probability
$$ \begin{aligned} & =P(X<2)=P(X=0)+P(X=1) \\ & ={ }^{50} C _0 \frac{4}{5}{ }^{50}+{ }^{50} C _1 \frac{4^{49}}{(5)^{50}}=\frac{4}{5}{ }^{49} \frac{4}{5}+\frac{50}{5}=\frac{54}{5} \frac{4}{5}{ }^{49} \end{aligned} $$
