Question

Four defective oranges are accidentally mixed with sixteen good ones. Three oranges are drawn from the mixed lot. The probability distribution of defective oranges is

• A. \(\begin{array}{ccccc} \mathrm{X} & 0 & 1 & 2 & 3 \\ \mathrm{P}(\mathrm{X}=x) & \frac{28}{57} & \frac{8}{95} & \frac{8}{19} & \frac{1}{285} \end{array}\)
• B. \(\begin{array}{ccccc} \mathrm{X} & 0 & 1 & 2 & 3 \\ \mathrm{P}(\mathrm{X}=x) & \frac{28}{57} & \frac{8}{19} & \frac{8}{95} & \frac{1}{285} \end{array}\)
• C. \(\begin{array}{ccccc} \mathrm{X} & 0 & 1 & 2 & 3 \\ \mathrm{P}(\mathrm{X}=x) & \frac{28}{57} & \frac{8}{95} & \frac{1}{285} & \frac{8}{19} \end{array}\)
• D. \(\begin{array}{ccccc} \mathrm{X} & 0 & 1 & 2 & 3 \\ \mathrm{P}(\mathrm{X}=x) & \frac{1}{285} & \frac{8}{95} & \frac{8}{19} & \frac{28}{57} \end{array}\)

Hint:

Use combinations: choose the defective and good oranges from the lot of twenty and divide by the ways of choosing three.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
There are 4 defective and 16 good oranges, 20 in all. Three are drawn, so the total number of ways is \(\binom{20}{3}=1140\).
Let X be the number of defective oranges drawn.

Step 2: Calculation for X = 0 and 1
\(P(X=0)=\dfrac{\binom{16}{3}}{1140}=\dfrac{560}{1140}=\dfrac{28}{57}\)
\(P(X=1)=\dfrac{4\cdot\binom{16}{2}}{1140}=\dfrac{480}{1140}=\dfrac{8}{19}\)

Step 3: Calculation for X = 2 and 3
\(P(X=2)=\dfrac{\binom42\cdot16}{1140}=\dfrac{96}{1140}=\dfrac{8}{95}\)
\(P(X=3)=\dfrac{\binom43}{1140}=\dfrac{4}{1140}=\dfrac{1}{285}\)

Step 4: Compare the options
The values \(\dfrac{28}{57},\dfrac{8}{19},\dfrac{8}{95},\dfrac{1}{285}\) in this order match option (B).

Final Answer:
The probabilities for 0, 1, 2 and 3 defective oranges are 28/57, 8/19, 8/95 and 1/285. \[ \boxed{\text{(B) }\text{P(X) = }\dfrac{28}{57},\ \dfrac{8}{19},\ \dfrac{8}{95},\ \dfrac{1}{285}} \]