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. 1/5
• B. 2/3
• C. 2/5
• D. 7/9
• E. 1/3

Hint:

When counting favorable outcomes for stock satisfaction, list all possible demand tuples that do not exceed stock, then count the number of permutations for each tuple.

Answer 1

The Correct Option is B

Step 1: Total possible color choices for 3 customers: each has 3 choices, so $3^3 = 27$ equally likely color preference combinations.
Step 2: Stock available: White: 2, Black: 2, Red: 1.
Step 3: We need the number of preference combinations that can be satisfied with the stock.
Step 4: Let $w, b, r$ be the number of customers wanting white, black, red respectively. $w+b+r=3$, $0 \le w \le 2$, $0 \le b \le 2$, $0 \le r \le 1$.
Step 5: List all possible $(w,b,r)$ tuples that satisfy constraints: $(2,1,0)$, $(2,0,1)$, $(1,2,0)$, $(0,2,1)$, $(1,1,1)$.
Step 6: Count number of sequences for each tuple: $(2,1,0)$: $\frac{3!}{2!1!0!} = 3$. $(2,0,1)$: $\frac{3!}{2!0!1!} = 3$. $(1,2,0)$: $\frac{3!}{1!2!0!} = 3$. $(0,2,1)$: $\frac{3!}{0!2!1!} = 3$. $(1,1,1)$: $\frac{3!}{1!1!1!} = 6$.
Step 7: Total favorable = $3+3+3+3+6 = 18$.
Step 8: Probability = $\frac{18}{27} = \frac{2}{3}$.
Step 9: Final Answer: The probability is $\frac{2}{3}$.