Question

A lot contains 50 defective & 50 non defective bulbs. Two bulbs are drawn at random, one at a time, with replacement. The ecents A, B, C are defined as
A= {the fist bulb is defective};
B= {the second bulb is non defective}
C= {the two bulbs are both defective or both non defective}
Determine whether A, B, C are independent.
If yes enter 1 else enter 0.

Answer 1

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.
$A=$ {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 } $=\{D_1{D}_2,N_1{N}_2\}$
Again, we know that
$A\cap B=\left\{D_1{N}_2\right\},B\cap C=\left\{N_1{N}_2\right\}$
$C\cap A=\left\{D_1{D}_2\right\}$ and $A\cap B\cap C=\varphi$
Also,$P\left(A\right)=P\left\{D_1{D}_2\right\}+P\left\{D_1{D}_2\right\}=P\left(D_1\right)P\left(D_2\right)+P\left(D_1\right)P\left(N_2\right)$
$=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{2}$
Similarly, $P\left(B\right)=\frac{1}{2}$ and $P\left(C\right)=\frac{1}{2}$
Also, $p(A\cap B)=P\left(D_1{N}_2\right)=P\left(D_1\right)P\left(N_2\right)=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}$
Similarly $P(B\cap C)=\frac{1}{4},P(C\cap A)=\frac{1}{4}$ and $P(A\cap B\cap C)=0$
Since, $P(A\cap B)=P\left(A\right)P\left(B\right),P(B\cap C)=P\left(B\right)P\left(C\right)$ and $P(C\cap A)=P\left(C\right)P\left(A\right)$
Therefore $A,B$ and $C$ are pairwise independent.
Also $P(A\cap B\cap C)\ne P\left(A\right)P\left(B\right)P\left(C\right)$ therefore $A,B$ and $C$ cannot be independent