SATHEE: Probability 3 Question 39

Probability 3 Question 39

39. 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).

Determine whether

(i) $A, B, C$ are pairwise independent.

(ii) $A, B, C$ are independent.

(1992, 6M)

Show Answer

Answer:

Correct Answer: 39. (i) $A, B$ and $C$ are pairwise independent

Solution:

Let $D_{1}$ denotes the occurrence of a defective bulb in Ist draw.

Therefore, $P (D_{1}) = \frac{50}{100} = \frac{1}{2}$

and let $D_{2}$ denotes the occurrence of a defective bulb in IInd draw.

Therefore, $P (D_{2}) = \frac{50}{100} = \frac{1}{2}$

and let $N_{1}$ denotes the occurrence of non-defective bulb in Ist draw.

Therefore, $P (N_{1}) = \frac{50}{100} = \frac{1}{2}$

Again, let $N_{2}$ denotes the occurrence of non-defective bulb in IInd draw.

Therefore, $P (N_{2}) = \frac{50}{100} = \frac{1}{2}$

Now, $D_{1}$ is independent with $N_{1}$ and $D_{2}$ is independent with $N_{2}$ .

According to the given condition,

$A = $ t h e f i r s t b u l b i s d e f e c t i v e $ = D_{1} D_{2}, D_{1} N_{2}$

$B = $ t h e s e c o n d b u l b i s n o n - d e f e c t i v e $ = D_{1} N_{2}, N_{1} N_{2}$

and $C = $ t h e t w o b u l b s a r e b o t h d e f e c t i v e $$

$= D_{1} D_{2}, N_{1} N_{2}$

Again, we know that,

$\begin{aligned} A \cap B = D_{1} N_{2}, B \cap C = N_{1} N_{2} . \\ C \cap A = D_{1} D_{2} and A \cap B \cap C = φ \end{aligned}$

Also,

$\begin{aligned} P (A) & = P D_{1} D_{2} + P D_{1} N_{2} \\ = P (D_{1}) P (D_{2}) + P (D_{1}) P (N_{2}) \\ = \frac{1}{2} \frac{1}{2} + \frac{1}{2} \frac{1}{2} = \frac{1}{2} \end{aligned}$

Similarly, $P (B) = \frac{1}{2}$ and $P (C) = \frac{1}{2}$

Also, $P (A \cap B) = P (D_{1} N_{2}) = P (D_{1}) P (N_{2}) = \frac{1}{2} \frac{1}{2} = \frac{1}{4}$ Similarly, $P (B \cap C) = \frac{1}{4}, P (C \cap A) = \frac{1}{4}$