JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Probability

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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 :

(A) $\frac{164}{25} (\frac{1}{5})^{48}$
(B) $\frac{201}{5} (\frac{1}{5})^{49}$
(C) $\frac{54}{5} (\frac{4}{5})^{49}$ ✓ Correct
(D) $\frac{316}{25} (\frac{4}{5})^{48}$

Answer: (C)

Solution

Failing to solve fewer than two means he misses

0

1

problem, with the number missed following

Bin (50, \frac{1}{5})

. So the probability is

(\frac{4}{5})^{50} + (1 50) (\frac{4}{5})^{49} \frac{1}{5} = (\frac{4}{5})^{49} (\frac{4}{5} + 10) = \frac{54}{5} (\frac{4}{5})^{49}

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option

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