Question 1027044: in a sample of 1200 adults, 200 dine out more than once a week. 2 adults are selected at random without replacement. Assuming the sample is representative of all adults, what is the probability that both adults dine out more than once a week?
Answer by mathmate(429)   (Show Source):
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Question:
in a sample of 1200 adults, 200 dine out more than once a week. 2 adults are selected at random without replacement. Assuming the sample is representative of all adults, what is the probability that both adults dine out more than once a week?
Solution:
The question requests us to assume that the two adults sample are representative of the population.
So P(dine-out)=200/1200=1/6
For a two step experiment each with a probability of 1/6, the probability of both events succeed is the product, by the multiplication rule, giving
P(both dine-out)=(1/6)(1/6)=1/36=0.02778 (approx.)
In reality, once the first has been sampled, there leaves 199 out-diners out of the 1199 remaining people, so the actual probability is
P(both dine-out)=(1/6)*(199/1199)=199/7194=0.02766 approx.
The last result can also be obtained using the hypergeometric distribution, which is
P(2 successes out of 120)
=C(200,2)C(1000,0)/(1200,2)
=199/7194 as before.
When the population is large, and the probability of success is not extremely low, the specified approximation (of constant probability) is quite close to the exact value, as illustrated by the example.
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