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: Understanding the Question:
There are \(8\) batteries, \(3\) defective and \(5\) good. Two are drawn, so \(X\in\{0,1,2\}\). Total ways: \(\binom82 = 28\).

Step 2: Probabilities:
\(P(X=0) = \frac{\binom52}{28} = \frac{10}{28}\).
\(P(X=1) = \frac{\binom31\binom51}{28} = \frac{15}{28}\).
\(P(X=2) = \frac{\binom32}{28} = \frac{3}{28}\).
The total is \(\frac{10+15+3}{28} = 1\).

Step 3: Match:
This is the table in option A. Options B and D start at \(X = 1\), which cannot be (the number of defective ones can be \(0\)). Option C swaps \(\frac{10}{28}\) and \(\frac{15}{28}\).

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