Combinations problem with 'at least' condition

Question

Question: Suppose that in a group of 20 people, 5 have brown hair. If we randomly select 4 individuals from this group of 20, what is the probability that at least 1 has brown hair?

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My attempt:

Intuitively I know that this probability can be found by just doing this:

P(at least one with brown hair) = 1 – P(no one has brown hair)

$= 1 - [(5/20) * (5/19) * (5/18) * (5/17)] $

$= 1 – 0.005374957$

$= 0.995$ (to 3 d.p.)

However, I'm trying to instead use a combinations method to solve the problem which I'm not quite as familiar with. I'm not really sure how to approach it by I'm guessing the answer has $20\choose4$ in the denominator?

Answer 1

Let's correct your attempt, then address how to solve the problem using combinations.

It is true that $$\Pr(\text{at least one with brown hair}) = 1 - \Pr(\text{none with brown hair})$$

Since $5$ of the $20$ people have brown hair, $20 - 5 = 15$ do not have brown hair. Thus, the probability that the first person selected does not have brown hair is $15/20$. If the first person selected does not have brown hair, the probability that the second person selected also does not have brown hair is $14/19$ since $14$ of the $19$ individuals who are left do not have brown hair. If the first two people selected do not have brown hair, the probability that the third person selected also does not have brown hair is $13/18$. If the first three people selected do not have brown hair, the probability that the fourth person selected also does not have brown hair is $12/17$. Hence, $$\Pr(\text{none with brown hair}) = \frac{15}{20} \cdot \frac{14}{19} \cdot \frac{13}{18} \cdot \frac{12}{17}$$ so $$\Pr(\text{at least one with brown hair}) = 1 - \frac{15}{20} \cdot \frac{14}{19} \cdot \frac{13}{18} \cdot \frac{12}{17}$$

Using combinations, the probability that none of the four selected people have brown hair is the number of ways of selecting four of the fifteen people who do not have brown hair divided by the number of ways of selecting four of the twenty people, which is $$\Pr(\text{none with brown hair}) = \frac{\dbinom{15}{4}}{\dbinom{20}{4}}$$ Hence, $$\Pr(\text{at least one with brown hair}) = 1 - \frac{\dbinom{15}{4}}{\dbinom{20}{4}}$$

Answer 2

Let’s see how many combinations we have: $1.$ $1$ brown hair in the group. $2.$ $2$ brown hair in the group. $3.$ $3$ brown hair in the group. $4.$ $4$ brown hair in the group. If there’s $1$ brown hair , it would be : $P(5;1,4) * P(15;3,11)$. If there are $2$ brown hair , it would be : $P(5;2,3) * P(15;3,12)$. If there are $3$ brown hair , it would be : $P(5;3,2) * P(15;1,14)$. And if there are 4 brown hair in the group, there would be any room left for the others so that’s a $P(5;4,1)$. And then , you have to divide the answer by the number of ways we can pick 4 people from 20 people which would be: $P(20;4,16)$ I hope this helps!