Question

1. There are three brands, say X, Y and Z of an item available in the market. A consumer chooses exactly one of them for his use. He never buys two or more brands simultaneously. The probabilities that he buys brands X, Y and Z are 0.20, 0.16 and 0.45, respectively.

a) What is the probability that he does not buy any of the brands?

b) Given that a customer buys some brand, what is the probability that he buys brand

Answer 1

Given

$$P(X\cap Y)=P(X\cap Z)=P(Y\cap Z)=P(X\cap Y\cap Z)=0$$

Then

$$P(X\cup Y\cup Z)=P(X)+P(Y)+P(Z)$$

$$-P(X\cap Y)-P(X\cap Z)-P(Y\cap Z)$$

$$+P(X\cap Y\cap Z)$$

$$=P(X)+P(Y)+P(Z)-0-0-0+0$$

$$=P(X)+P(Y)+P(Z)$$

a)

$$P(X'\cap Y'\cap Z')=1-P(X\cup Y\cup Z)$$

$$=1-(P(X)+P(Y)+P(Z))$$

$$=1-(0.20+0.16+0.45)=0.19$$

b)

$$P(X|(X\cup Y\cup Z))=\dfrac{P(X\cap(X\cup Y\cup Z))}{P(X\cup Y\cup Z)}$$

$$=\dfrac{P(X)}{P(X)+P(Y)+P(Z)}$$

$$=\dfrac{0.20}{0.20+0.16+0.45} =\dfrac{20}{81}\approx0.2469$$