A consumer organization estimates that 32% of the households in a particular community have one television set, 38% have two sets, and 16% have three or more sets. If two households are chosen at random, what is the probability that at least one has one or more television sets

Question

Iris Riley

Mathematics

13 October, 03:10

A consumer organization estimates that 32% of the households in a particular community have one television set, 38% have two sets, and 16% have three or more sets. If two households are chosen at random, what is the probability that at least one has one or more television sets

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Shea Best · Answer 1

If two households are chosen at random, the probability that at least one has one or more television sets is equal to 0.9804 or (98.04%)

Step-by-step explanation:

32% of the households have one television set, 38% have two sets 16% have three or more sets

Then 100% - (32% + 38% + 16%) = 14% of households does not have any television set

the probability that both of the two randomly chosen households does not have any television set is 0.14*0.14 = 0.0196

If two households are chosen at random, the probability that at least one has one or more television sets is equal to

1 - (the probability that both of the two randomly chosen households does not have any television set) = 1 - 0.0196 = 0.9804 (98.04%)