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The proportion of the American adult population that supports candidate Green is p=0.22. A SRS of 9 adults asks if they agree with the statement “I support candidate Green.” What is the probability that at least 2 of those surveyed would agree with that statement?

Ok, so far I know that the probability of at least 1 person would be .8931 -edited- was using .88 instead .78- Is that useful in solving for at least two?

Evan Marsh
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2 Answers2

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The probability that at least $2$ support Green is $1$ minus the probability that $0$ people or $1$ person supports Green.

The probability that no one supports Green is $(0.78)^9$.

The probability that exactly $1$ person supports Green is $\dbinom{9}{1}(0.22)(0.78)^8$.

We could also do the problem the hard way. The probability that exactly $k$ people support Green is $\dbinom{9}{k}(0.22)^k(0.78)^{9-k}$. Calculate all of these, for $k=2$ to $9$, and add up.

Remark: We explain why the probability there is exactly one Green supporter is $9(0.22)(0.78)^8$. Line up the people we are interviewing. The probability that the first is a supporter of Green and the rest are not is $(0.22)(0.78)^8$. The probability that the first does not support Green, but the second does, and the rest don't is $(0.78)(0.22)(0.78)^7$, which is $(0.22)(0.78)^8$. Similarly, the probability the first two are not supporters, but the third is a supporter, and the rest are not is $(0.22)(0.78)^8$. We continue this way through all $9$ people, and end up with $(0.22)(0.78)^8$ added to itself $9$ times, for a total of $(9)(0.22)(0.78)^8$. The point is that there are $9$ different ways in which there is $1$ supporter of Green and $8$ non-supporters.

André Nicolas
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  • Sorry for the ignorance, but when you have 9 over k in the parentheses, what is that? – Evan Marsh Mar 02 '13 at 00:37
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    It is the binomial coefficient. The symbol $\binom{n}{r}$ represents the number of ways to choose $r$ objects from a collection of $n$ objects. In high school, it is sometimes called $C(n,r)$, or $C_r^n$. So $\binom{9}{1}=9$. – André Nicolas Mar 02 '13 at 00:42
  • So then the answer would be around 86%, right? Just using the shorter way I calculated 1-(P(0)+P(exactly 1) – Evan Marsh Mar 02 '13 at 00:51
  • I get that $(0.78)^9\approx 0.1068$ and $9(0.22)(0.78)^8\approx 0.2713$. The sum is $\approx 0.378$. Take away from $1$. I get about $0.622$. Haven't figured out where your calculation went astray. Maybe mine also did, but your number is for sure much too large. – André Nicolas Mar 02 '13 at 01:01
  • Hey sorry for taking so long, but why did you multiply by 9? Is the coefficient effectively just a fraction? If so, that is where I went wrong. – Evan Marsh Mar 02 '13 at 02:29
  • @EvanMarsh: I will add an explanation of this special case in a remark at the end, it will take a few minutes. – André Nicolas Mar 02 '13 at 02:34
  • Your remark makes sense, but would this way also work?: 1- (P(0)+P(1)). That is what I did, arriving at .8629 – Evan Marsh Mar 02 '13 at 02:56
  • You are calculating what you call $P(1)$ incorrectly. I used exactly the same expression $1-(P(0)+P(1))$ as you did. To repeat, $P(1)=9(0.22)(0.78)^8$. – André Nicolas Mar 02 '13 at 03:14
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$$ \mathcal P( \text{at least two agree}) = 1 - \mathcal P(\text{ eight or nine disagree}) = 1- {9 \choose 8}p^8(1-p) - {9 \choose 9} p^9 = 1- p^8(9-8p),\quad p = .78$$

will.fiset
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Mohamad
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