Question

It is known that a box of \(8\) batteries contains \(3\) defective pieces and a person randomly selects \(2\) batteries from this box. Then the probability distribution of the number of defective batteries is

• A.

  \(X = x\)	\(0\)	\(1\)	\(2\)
  \(P(X = x)\)	\(\frac{10}{28}\)	\(\frac{15}{28}\)	\(\frac{3}{28}\)

• B.

  \(X = x\)	\(1\)	\(2\)	\(3\)
  \(P(X = x)\)	\(\frac{10}{28}\)	\(\frac{15}{28}\)	\(\frac{3}{28}\)

• C.

  \(X = x\)	\(0\)	\(1\)	\(2\)
  \(P(X = x)\)	\(\frac{15}{28}\)	\(\frac{10}{28}\)	\(\frac{3}{28}\)

• D.

  \(X = x\)	\(1\)	\(2\)	\(3\)
  \(P(X = x)\)	\(\frac{15}{28}\)	\(\frac{10}{28}\)	\(\frac{3}{28}\)

Hint:

\(X\) can be 0, 1 or 2, and the probabilities use combinations out of \(\binom82=28\).

Answer 1

The Correct Option is A

Step 1: Plan:
Use the sequential method of drawing without replacement.

Step 2: Steps:
$P(0) = \frac58\cdot\frac47 = \frac{20}{56} = \frac{10}{28}$. $P(2) = \frac38\cdot\frac27 = \frac{6}{56} = \frac{3}{28}$.
$P(1) = 1 - \frac{10}{28} - \frac{3}{28} = \frac{15}{28}$. So the pairs (0, 10/28), (1, 15/28), (2, 3/28) match option A.

Final Answer:
The distribution is $P(0)=\frac{10}{28}$, $P(1)=\frac{15}{28}$, $P(2)=\frac{3}{28}$, option (A). \[ \boxed{\text{Option (A)}} \]