Probability Ques 120
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) $\frac{201}{5} (\frac{1}{5})^{49}$
(b) $\frac{316}{25} (\frac{4}{5})^{48}$
(c) $\frac{54}{5} (\frac{4}{5})^{49}$
(d) $\frac{164}{25} (\frac{1}{5})^{48}$
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Answer:
Correct Answer: 120.(c)
Solution:
Formula:
- 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} $
BETA
