Question

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 three apply for the same house is:

• A. $\frac{2}{9}$
• B. $\frac{1}{9}$
• C. $\frac{4}{9}$
• D. $\frac{1}{27}$

Hint:

Alternatively: the first person can choose any house (prob = 1). The second and third persons must choose that same house, each with probability $1/3$. Total probability $= 1 \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$.

Answer 1

The Correct Option is B

Step 1: Set up the counting.
Each person freely picks one of the 3 houses. Probability is favourable choices over all choices.

Step 2: Count all possibilities.
Each of 3 persons has 3 options, so \[ n(S) = 3 \times 3 \times 3 = 27 \]

Step 3: Describe the good event.
We want all three to pick the same house.

Step 4: Count the good ways.
They could all pick house 1, or all house 2, or all house 3, so $n(E) = 3$.

Step 5: Form the probability.
\[ P = \frac{3}{27} \]

Step 6: Simplify.
\[ P = \frac{1}{9} \] \[ \boxed{ \frac{1}{9} } \]