Question

let x be the number of smokers in a family composed of a husband and a wife. X is a random variable which takes values 0,1,2, with respective probabilities po, p1, p2. The average number of smokers per family is 0.63. 40% of families contain at least one smoker. find the probabilities po, p1, p2.

Note: this question is related to binomial distribution lesson  

Answer 1

Here, it is not given in the question that the distribution is Binomial so we don't make any assumption about the distribution of X.

Now, we are given that X (the number of smokers in a family) is a random variable which takes values 0,1 and 2 with respective probabilities p0,p1 and p2.

Thus, p0 + p1 + p2 = 1..................................(1)

Now, we are given that average number of smokers per family is 0.63. Thus:

Moreover, we are given that 40% of families contain at least one smoker. Thus:

Subtracting equation (3) from equation (1), we get:

p0 + p1 + p2 - [p1 + p2] = 1 - 0.40

=> p0 = 0.60 [Answer]

Subtracting equation (3) from equation (2), we get:

p1 + 2*p2 - [p1 + p2] = 0.63 - 0.40

=> p2 = 0.23 [Answer]

Substituting the value of p2 in equation (3), we get:

p1 + 0.23 = 0.40

=> p1 = 0.17 [Answer]

Thus,

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