Probability Calculation When The Sample Is Given As A Ratio?

A friend of mine asked me to solve a problem for her, but I didn't know how to start, can someone please explain how I would go about solving this problem? Here's the problem:

The ratio of males to female in a room is 9:11. A third of the men have black hair, and half of the women have black hair and the rest of them have blonde hair.

8 people are chosen at random.

a). What is the probability that these eight people are all male?

b). What is the probability that there are exactly five females?

c). What is the probability that there are at least two black-haired people?

I thought that it was an easy question until I got to the last part and then I started to question whether or not I got the rest of it right. Please let me know how to solve it so that I can do that.

The most annoying thing about this question is the ratio.

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

In order to obtain a precise answer, any random sample that is done without replacement must require knowledge of the population size from which the sample is drawn. As the question stands, the most we can say about the population is that there are 18n males and 22n females, where n is some positive integer (which is the most general condition that satisfies the given criteria). If n is sufficiently large (i.e., so that 40n is much greater than 8), then we can approximate the desired probabilities by considering the sampling of people as occurring with replacement.

In terms of n, the answer to part (a) is easily expressed as a probability mass function for a hypergeometric distribution: $\frac{\binom{18n}{8} \binom{22n}{0}}{\binom{40n}{8}},$ the limit of which as n tends to infinity is simply the binomial probability $(9/20)^8$ .
The answer to part (b) is $\frac{\binom{18n}{3} \binom{22n}{5}}{\binom{40n}{8}},$ and again, the limit of this as n tends to infinity is $\binom{8}{3} (9/20)^3 (11/20)^5$ .
For part (c), we observe that there are $6n + 11n = 17n$ black-haired people in a population of size $40n$ , so the desired probability is $1 - \frac{\binom{17n}{0} \binom{23n}{8}}{\binom{40n}{8}} - \frac{\binom{17n}{1} \binom{23n}{7}}{\binom{40n}{8}},$ and the limit of this is again a binomial probability, but this time with probability of success $p = 17/40$ : $1 - \binom{8}{0}(17/40)^0 (23/40)^8 - \binom{8}{1} (17/40)^1 (23/40)^7.$

hero p. - 8 years, 2 months ago

You could have also got this answer from the formula for bernouilli's trials, although this way also the answer is correct.

Pratik Singhal - 8 years, 2 months ago

As I pointed out at the very beginning of my post, you cannot precisely model the desired outcomes if the number of people in the room is finite, because the question as stated implies sampling without replacement. Bernoulli trials do not apply except in the limiting case, and even then, what we are interested in is the number of successes, which is modeled as a binomial distribution (which arises from considering the sum of a fixed number of IID Bernoulli trials).

The whole point of my previous response was to explain that no precise numerical probabilities can be given without knowing the population size--using a binomial distribution is an approximation under the assumption that there are many people in the room. It would not hold if n is small.

hero p. - 8 years, 2 months ago

When facing these kinds of problems you can do an empirical approach based on the fact that as you scale things up or down the probabilities remain the same (logically if they don't tell you the sample size is because this assumption is true, though you can prove this). For example, you can take 18 men and 22 women (we must divide by 2 and we don't want 4 and a half women :p ). From these 18/3=6 men have black hair and 22/2=11 women have black hair. In total we have 17 B's (Black Hairs) and 23 Y's (Blondes).

Now we have to choose 2,3,4,5,6,7 and 8 B's, but its easier to count the number of cases we dont have to count, that is 0 and 1 B's. To choose 0 B's is to choose 8 out of 23 or 23C8. To choose 1 B is to choose 1 out of 17 and 7 out of 23 or (17C1)x(23C7). We must add all these cases and substract them from the total (choose 8 out of 40 people or 40C8 ).

The probability is just the number of cases over the total amount of cases or: $P=\frac {{40 \choose 8} - {23 \choose 8}-{23 \choose 7} \times {17 \choose 1}}{{40 \choose 8}}=1 - \frac {{23 \choose 8} + {23 \choose 7} \times {17 \choose 1}}{{40 \choose 8}}$

Sebastian Garrido - 8 years, 2 months ago

I don't know how to size up my latex. Anyone?

Sebastian Garrido - 8 years, 2 months ago

There is not enough information to solve (e.g. for part a the answer if there are 9 men and 11 women would be 9 / (11C8) whereas the answer for part a if there are 18 men and 22 women would be (18C8) / (40C8) which when evaluated give different results.

EDIT. Whoops, slight mistake on the second case. Sorry!

Francisco Rivera - 8 years, 2 months ago

I believe your second case should be 22 women? That would explain the down votes.

Calvin Lin Staff - 8 years, 2 months ago

Francisco, I think you got the problem a wrong, the solution should not be line that, instead you should consider this problem as a problem of bernoulli's trials. I got the right answer after solving it from bernouilli's trials.

Pratik Singhal - 8 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$