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Probability question

2005 · Shift 0 · Q87

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2005 · Shift 0 · Q87

Probability question

2005 · Shift 0 · Q87

JEE MainMathematicsProbabilityMCQ+4 / −1

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 :

A
${2 \over 9}$
B
${1 \over 9}$
C
${8 \over 9}$
D
${7 \over 9}$

View written solutionFree

Correct answer: B

Total number of possible application patterns

Each of the 3 persons independently applies for 1 of the 3 houses.

So, for each person, there are 3 choices. Hence total outcomes:

$3^3 = 27$

Favourable outcomes: all three apply for the same house

This can happen in exactly 3 ways:

all apply for house 1
all apply for house 2
all apply for house 3

So, favourable outcomes = $3$ .

Required probability

$P(\text{all three apply for the same house}) = \frac{3}{27} = \frac{1}{9}$

Check options

A: $\frac{2}{9}$ — incorrect
B: $\frac{1}{9}$ — correct
C: $\frac{8}{9}$ — incorrect
D: $\frac{7}{9}$ — incorrect

Therefore, the correct answer is:

$\boxed{\frac{1}{9}}$

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