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(D) There are $3$ houses and $3$ persons. Each person can choose any of the $3$ houses independently.Total number of ways for $3$ persons to choose ho

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$\frac{1}{9}$

Vedclass · Answer

(D) There are $3$ houses and $3$ persons. Each person can choose any of the $3$ houses independently.Total number of ways for $3$ persons to choose houses $= 3 	imes 3 	imes 3 = 27$.For all $3$ persons to apply for the same house,they must all choose house $1$,or all choose house $2$,or all choose house $3$.Number of favorable outcomes $= 3$ (i.e.,{House $1, 1, 1$},{House $2, 2, 2$},{House $3, 3, 3$}).Probability $= \frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{3}{27} = \frac{1}{9}$.

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is

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