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)
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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, $P\left(D _1\right)=\frac{50}{100}=\frac{1}{2}$
and let $D _2$ denotes the occurrence of a defective bulb in IInd draw.
Therefore, $P\left(D _2\right)=\frac{50}{100}=\frac{1}{2}$
and let $N _1$ denotes the occurrence of non-defective bulb in Ist draw.
Therefore, $P\left(N _1\right)=\frac{50}{100}=\frac{1}{2}$
Again, let $N _2$ denotes the occurrence of non-defective bulb in IInd draw.
Therefore, $P\left(N _2\right)=\frac{50}{100}=\frac{1}{2}$
Now, $D _1$ is independent with $N _1$ and $D _2$ is independent with $N _2$.
According to the given condition,
$A={$ the first bulb is defective $}={D _1 D _2, D _1 N _2 }$
$B={$ the second bulb is non-defective $}={D _1 N _2, N _1 N _2 }$
and $C={$ the two bulbs are both defective $}$
$$ ={D _1 D _2, N _1 N _2 } $$
Again, we know that,
$$ \begin{aligned} & A \cap B={D _1 N _2 }, B \cap C={N _1 N _2 } . \\ & C \cap A={D _1 D _2 } \text { and } A \cap B \cap C=\varphi \end{aligned} $$
Also,
$$ \begin{aligned} P(A) & =P{D _1 D _2 }+P{D _1 N _2 } \\ & =P\left(D _1\right) P\left(D _2\right)+P\left(D _1\right) P\left(N _2\right) \\ & =\frac{1}{2} \quad \frac{1}{2}+\frac{1}{2} \quad \frac{1}{2}=\frac{1}{2} \end{aligned} $$
Similarly, $P(B)=\frac{1}{2}$ and $P(C)=\frac{1}{2}$
Also, $\quad P(A \cap B)=P\left(D _1 N _2\right)=P\left(D _1\right) P\left(N _2\right)=\frac{1}{2} \quad \frac{1}{2}=\frac{1}{4}$ Similarly, $P(B \cap C)=\frac{1}{4}, P(C \cap A)=\frac{1}{4}$
and $P(A \cap B \cap C)=0$.
Since, $\quad P(A \cap B)=P(A) P(B), P(B \cap C)=P(B) P(C)$
and $\quad P(C \cap A)=P(C) P(A)$
Therefore, $A, B$ and $C$ are pairwise independent.
Also, $P(A \cap B \cap C) \neq P(A) P(B) P(C)$ therefore $A, B$ and $C$ cannot be independent.
