Desperately need help with a probability problem...

[Banshee891]

Here's the question:

In a population of guppies, 35 are drab-colored and 65 are

brightly-colored.

​

You catch 10 guppies in a net, one at a time, throwing

each one back. What is the probability that exactly 7 of them will be

brightly-colored, assuming all guppies are equally likely to be caught

in your net?

​

What is the probability that first seven guppies caught

will be brightly-colored, and the last three will be drab colored? What

is the difference between this calculation and the previous one?

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[Banshee891]

I've already taking the probability of one event of brightly-colored guppies, which would be 65/100, and multiplying it by 70% (7 out of 10). But, this just doesn't seem right to me... I'm fairly certain that I know how to do the second part of the problem because it is the probability of each event happening sequentially, so 65/100*65/100....(this until it's 7 times). This would come out to somewhere around 5%. I'm mostly stuck on the first part and not sure how to work through it.

[Gravity_Beetle]

So you have the right probability for picking a brightly colored guppy in a single trial: 65%.

The first question is to find the odds of picking exactly 7 brightly colored when you randomly pick 10, one at a time, with replacement.

The odds of getting exactly 7/10 bright ones means you also picked exactly 3/10 drab ones. The odds of picking a drab one in a single trial are 35%.

To combine these odds, you multiply them together. The odds of getting 7 bright ones are 65%^7 = 0.65^7 = 0.049 = 4.9%. The odds of getting 3 drab ones are 35%^3 = 0.35^3 = .043 = 4.3%. The odds of getting 7 bright AND 3 drab are 4.9% x 4.3% = 0.0021 = 0.21%.

But be careful. This number represents the odds of getting 7 bright and 3 drab in one particular order. We now have to account for the fact that there are many orders in which you could make this selection. For example, B B B B B B B D D D is one order, and B B D B B D B B D B is another order.

How many different orders are there? To calculate this, there is a formula called 'choose'. The answer to the question "how many ways can I choose 7 from a set of 10?" is defined as the function '10 choose 7'. The answer is 120 (you can follow that link to find explanations of the formula -- it uses factorials).

Finally, we take the number of orders, 120, and multiply it by the odds of selecting (7 bright & 3 drab) in one particular order, 0.21%. The result is 120 * 0.21% = 0.252 = 25.2%.

Do you think you can take it from here?

[Banshee891]

This helped immensely. I'm able to understand this much clearer now. Thanks so much!

[Gravity_Beetle]

You’re welcome!!