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 $A,{\text{ }}B,{\text{ }}C$ 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? $\left( 1 \right)$ $A,{\text{ }}B,{\text{ }}C$ are pairwise independent. $\left( 2 \right)$ $A,{\text{ }}B,{\text{ }}C$ are independent.
A. Only $\left( 1 \right)$ is true.
B. Both $\left( 1 \right)$ and $\left( 2 \right)$ are true.
C. Only $\left( 2 \right)$ is true.
D. Both $\left( 1 \right)$ and $\left( 2 \right)$ are false.

Answer 1

To solve this question, the concept of independent events is used. In probability, two events are independent if the occurrence of one event does not affect the occurrence of the other. The probability of occurring of the two events are independent of each other.

Complete step-by-step answer:
In the question, it is given that two bulbs are drawn at random, one at a time, with replacement so that means that there will be two draw counts.
Let the occurrence of a defective bulb In ${1^{st}}$ draw be ${D_1}$ and the probability of occurrence of a defective bulb in ${1^{st}}$ draw be $P({D_1})$ .
Therefore, $P({D_1}) = \dfrac{{50}}{{100}} = \dfrac{1}{2}$
Let the occurrence of a defective bulb In ${2^{nd}}$ draw be ${D_2}$ and the probability of occurrence of a defective bulb in ${2^{nd}}$ draw be $P({D_2})$ .
Therefore, $P({D_2}) = \dfrac{{50}}{{100}} = \dfrac{1}{2}$
And, let the occurrence of a non-defective bulb In ${1^{st}}$ draw be ${N_1}$ and the probability of occurrence of a non-defective bulb in ${1^{st}}$ draw be $P({N_1})$ .
Therefore, $P({N_1}) = \dfrac{{50}}{{100}} = \dfrac{1}{2}$
Again, let the occurrence of a non-defective bulb In ${2^{nd}}$ draw be ${N_2}$ and the probability of occurrence of a non-defective bulb in ${2^{nd}}$ draw be $P({N_2})$ .
Therefore, $P({N_2}) = \dfrac{{50}}{{100}} = \dfrac{1}{2}$
Now, ${D_1}$ is independent form ${N_1}$ and ${D_2}$ is independent from ${N_2}$ with the conditions shown below:
$A =$ {the first bulb is defective} $= \\{ {D_1}{D_2},{D_1}{N_2}\\}$
$B =$ {the second bulb is non defective} $= \\{ {D_1}{N_2},{N_1}{N_2}\\}$
$C =$ {the two bulbs are both defective or both non defective} $= \\{ {D_1}{D_2},{N_1}{N_2}\\}$
Therefore, we get that
$A \cap B = \\{ {D_1}{N_2}\\}$ , $B \cap C = \\{ {N_1}{N_2}\\}$ , $C \cap A = \\{ {D_1}{D_2}\\}$
$A \cap B \cap C = \emptyset$
Now, $P(A) = P\\{ {D_1}{D_2}\\} + P\\{ {D_1}{N_2}\\} = P({D_1})P({D_2}) + P({D_1})P({N_1}) = \dfrac{1}{2}.\dfrac{1}{2} + \dfrac{1}{2}.\dfrac{1}{2} = \dfrac{1}{2}$
Similarly, $P(B) = \dfrac{1}{2}$ , $P(C) = \dfrac{1}{2}$
Also, $P(A \cap B) = P({D_1}{N_2}) = P({D_1}).P({N_2}) = \dfrac{1}{2}.\dfrac{1}{2} = \dfrac{1}{4}$
Similarly, $P(B \cap C) = \dfrac{1}{4}$ , $P(C \cap A) = \dfrac{1}{4}$ , $P(A \cap B \cap C) = 0$
Since, we can see that $P(A \cap B) = P(A).P(B)$ , $P(B \cap C) = P(B).P(C)$ , $P(C \cap A) = P(C).P(A)$
Therefore, $A,{\text{ }}B,{\text{ }}C$ are pairwise independent.
We have $P(A \cap B \cap C) = 0$ and $P(A).P(B).P(C)$=$\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}=\dfrac{1}{8}$

Hence we can say that $P(A \cap B \cap C) \ne P(A).P(B).P(C)$
Therefore, $A,{\text{ }}B,{\text{ }}C$ are not independent.
Therefore, Only $\left( 1 \right)$ is true is the required answer.

So, the correct answer is “Option A”.

Note: An event $A$ is said to be independent of another event $B$ if the probability of occurrence of one of them is not affected by the occurrence of the other. For more clarity let us suppose, if we draw two cards from a pack of cards one after the other. The results of the two draws are independent if the cards are drawn with replacement i.e., the first card is put back into the pack before the second draw. If the cards are not replaced then the events of drawing the cards are not independent.