First slide

Introduction to probability

Question

Five different marbles are placed in 5 different boxes randomly. Then the probability that exactly two boxes remain empty is (each box can hold any number of marbles)

Moderate

A

$\frac{2}{5}$
B

$\frac{12}{25}$
C

$\frac{3}{5}$
D

none of these

Solution

We have,
$n (S) = 5^{5}$
For computing favorable outcomes, 2 boxes which are to remain empty, can be selected in ⁵C₂ ways and 5 marbles can be placed in the remaining 3 boxes in groups of 221 or 311 in
$3! [\frac{5!}{2! 2! 2!} + \frac{5!}{3! 2!}] = 150 ways$
$\Rightarrow n (A) =^{5} C_{2} \times 150$
Hence,
$P (E) =^{5} C_{2} \times \frac{150}{5^{5}} = \frac{60}{125} = \frac{12}{25}$

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