Question

A small store has five units of a new phone model in stock: two white, two black, and one red. Three customers arrive at the shop to buy a unit each. Each one has a pre- determined choice of the colour and will not buy a unit of any other colour. All the three customers are equally likely to have chosen any of the three colours. What is the probability that the store will be able to satisfy all the three customers?

• A. \( \frac{4}{5} \)
• B. \( \frac{7}{9} \)
• C. \( \frac{2}{3} \)
• D. \( \frac{8}{9} \)
• E. \( \frac{1}{3} \)

Answer 1

The Correct Option is C

Each customer chooses a colour independently and uniformly among {White, Black, Red}. So total possible outcomes for 3 customers = \(3^3 = 27\).

Step 1: Identify favourable cases. 

• The store has 2 white, 2 black, and 1 red.
• This means at most 2 customers can choose White, at most 2 customers can choose Black, and at most 1 customer can choose Red.

Step 2: Eliminate unfavourable cases.

• Case (i): All 3 choose White → Not possible (only 2 white). Count = 1 outcome.
• Case (ii): All 3 choose Black → Not possible (only 2 black). Count = 1 outcome.
• Case (iii): All 3 choose Red → Not possible (only 1 red). Count = 1 outcome.
• Case (iv): 2 Red + 1 other → Not possible (only 1 red). Count = \(\binom{3}{2} = 3\) outcomes.

Total unfavourable outcomes = \(1 + 1 + 1 + 3 = 6\).

Step 3: Favourable outcomes.

Favourable outcomes = \(27 - 6 = 21\).

Step 4: Probability.

\[ P = \frac{\text{Favourable}}{\text{Total}} = \frac{21}{27} = \frac{7}{9}. \]

Wait! Let us carefully verify again. The problem asks whether the store can serve all customers. That means each customer must get their chosen colour if possible.

Re-examining: If exactly 2 want White and 1 wants Black → possible. If exactly 2 want Black and 1 wants White → possible. But if 2 want Red → not possible. If all 3 want White or Black → not possible. If all 3 choose different → possible. If 2 want White, 1 Red → possible. If 2 want Black, 1 Red → possible. So correct count = 18 favourable out of 27.

\[ P = \frac{18}{27} = \frac{2}{3}. \]

Final Answer:

\(\boxed{\tfrac{2}{3}}\)