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Catracho Catracho
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5 years ago Edited: 5 years ago, bio_man
22
22​% of a certain​ country's voters think that it is too easy to vote in their country. You randomly select 12 likely voters. Find the probability that the number of likely voters who think that it is too easy to vote is​ (a) exactly​ three, (b) at least​ four, (c) less than eight.
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Educator
5 years ago
For this you use the binomial distribution formula: \(P\left(r\right)=_nC_r\times \pi ^r\left(1-\pi \right)^{n-r}\), where \(\pi\) is the probability.

22% is equivalent to 0.22 in decimal form.

a) For the probability of exactly 3, you set n = 12, r = 3, \(\pi\) = 0.22, and \(1-\pi =0.78\)

\(P\left(3\right)=_{12}C_3\times 0.22^3\left(0.78\right)^{12-3}\)

\(P\left(3\right)=_{12}C_3\times 0.22^3\left(0.78\right)^{12-3}=220\times 0.00113=0.250\)

b) For this, you have the option of finding the probability of 4 all the way to 12, or finding the sum of the probabilities of 0 to 3, and subtracting it from 1. I'll use the latter method:

\(P\left(r\ge 4\right)=1-\left[P\left(3\right)+P\left(2\right)+P\left(1\right)+P\left(0\right)\right]\)

\(1-\left[_{12}C_3\cdot 0.22^3\left(0.78\right)^9+_{12}C_2\cdot 0.22^2\left(0.78\right)^{10}+_{12}C_1\cdot 0.22^1\left(0.78\right)^{11}+_{12}C_0\cdot 0.22^0\left(0.78\right)^{12}\right]\)

\(P\left(r\ge 4\right)=0.261\)

c) For less than 8, find the probabilities of 8 to 12, then subtract from 1.

\(P\left(r<8\right)=1-\left[P\left(8\right)+P\left(9\right)+P\left(10\right)+P\left(11\right)+P\left(12\right)\right]\)

\(1-\left[_{12}C_8\cdot 0.22^8\left(0.78\right)^4+_{12}C_9\cdot 0.22^9\left(0.78\right)^3+_{12}C_{10}\cdot 0.22^{10}\left(0.78\right)^2+_{12}C_{11}\cdot 0.22^{11}\left(0.78\right)^1+_{12}C_{12}\cdot 0.22^{12}\left(0.78\right)^0\right]\)

\(P\left(r<8\right)=0.9989\rightarrow 0.999\)
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