Question

A lot contains $50$ defective and $50$ non-defective bulbs. Two bulbs are drawn at random, one at a time, with replacement. The events $A,B,C$ are defined as $A=$ {the first bulb is defective}, $B=$ {the second bulb is non defective}, $C=$ {the two bulbs are both defective or both non defective}, then which of the following statements is/are true?(1) $A,B,C$ are pair wise independent.(2) $A,B,C$ are independent.

• A. Only (1) is true.
• B. Both (1) and (2) are true.
• C. Only (2) is true.
• D. Both (1) and (2) are false.

Answer 1

The correct option is A Only (1) is true.

Let's denote the defective item by $D$ and non-defective by $\overline{D}$

$P\left(A\right)=P\left(D\right)\{P\left(\overline{D}\right)+P\left(D\right)\}=\frac{50}{100}\times 1=\frac{1}{2}$

$P\left(B\right)=\{P\left(\overline{D}\right)+P\left(D\right)\}\times P\left(\overline{D}\right)=1\times \frac{50}{100}=\frac{1}{2}$

$P\left(C\right)=P(DD\cup \overline{D}\overline{D})=P\left(DD\right)+P\left(\overline{D}\overline{D}\right)-P(DD\cap \overline{D}\overline{D})$

$\Rightarrow P\left(C\right)=\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2}-0=\frac{1}{2}$

Now

$P(A\cap B)=P\left(D\overline{D}\right)=\frac{50}{100}\times \frac{50}{100}=\frac{1}{4}$

$P(B\cap C)=P\left(\overline{D}\overline{D}\right)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

$P(C\cap A)=P\left(DD\right)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

Thus, we can see that

$P(A\cap B)=P\left(A\right)P\left(B\right),P(B\cap C)=P\left(B\right)P\left(C\right)$

$P(C\cap A)=P\left(C\right)P\left(A\right)$