People per Household

In the 2001 population census, information was collected about the number of people in each household, which is denoted by X X . It is given that P ( X = 1 ) = 0.25 P(X=1) = 0.25 , P ( X = 2 ) = 0.32 P(X=2) = 0.32 , P ( X = 3 ) = 0.18 P(X=3) = 0.18 , P ( X = 4 ) = 0.15 P(X=4) = 0.15 , P ( X = 5 ) = 0.07 P(X=5) = 0.07 , P ( X = 6 ) = 0.02 P(X=6) = 0.02 and P ( X 7 ) = 0.01 P(X\geq 7) = 0.01 . If R R is the probability that there are at most 4 4 people in the household, what is the value of 100 R 100R ?


The answer is 90.

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2 solutions

Alvin Willio
Aug 4, 2013

R = P1 + P2 + P3 + P4

R = 0.25 + 0.32 + 0.18 + 0.15

R = 0.90

100 R = 0.90 x 100

100 R = 90

Since the events are disjoint (as an example, a household cannot house 1 person and 2 persons at the same time), the probability that either of the events P ( X = 1 ) , P ( X = 2 ) , P ( X = 3 ) P(X=1), P(X=2), P(X=3) or P ( X = 4 ) P(X=4) happens, is simply the sum of their probabilities.

That is P ( X 4 ) = 0.25 + 0.32 + 0.18 + 0.15 = 0.90 P(X\leq 4)=0.25+0.32+0.18+0.15=0.90

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