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In a certain congressional district, it is known that 30% of the registered voters classify themselves as conservatives. If ten registered voters are selected at random from this district, what is the probability that two of them will be conservatives? (Round your answer to four decimal places.)

User FoJjen
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1 Answer

5 votes

Answer:

0.2334

Explanation:

We know that 30% are conservatives, so that is our proportion.

We define then,

p=0.30

n=0.7 (probability to not be conservative)

Our sample size (n) is 10.

So need to find if three of them are conservatives

This a typical binomial distribution


p(\pi)= ^(m)c_(\pi)p*n^(m-\pi)

Here
\pi=2, (you can be either conservative or not conservative)


p(2)=^(10)c_2*(0.3)^2(0.7)^(10-2)\\p(2)=(\frac{10*9}(2))*(0.3)*(0.7)^(10-2)\\p(2)=0.2334

User Riggaroo
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7.4k points