JEE PYQ: Probability Question 39
Question 39 - 2019 (12 Apr Shift 2)
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 problems is:
(1) $\frac{201}{5}\left(\frac{1}{5}\right)^{49}$
(2) $\frac{316}{25}\left(\frac{4}{5}\right)^{48}$
(3) $\frac{54}{5}\left(\frac{4}{5}\right)^{49}$
(4) $\frac{164}{25}\left(\frac{1}{5}\right)^{48}$
Show Answer
Answer: (3)
Solution
$P(\text{unable to solve} < 2) = P(\text{solve all 50}) + P(\text{unable to solve 1}) = {}^{50}C_0\left(\frac{4}{5}\right)^{50} + {}^{50}C_1\left(\frac{1}{5}\right)\left(\frac{4}{5}\right)^{49} = \left(\frac{4}{5}\right)^{49}\left[\frac{4}{5} + 10\right] = \frac{54}{5}\left(\frac{4}{5}\right)^{49}$.
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