Thread: Probability applications of counting principles (permutations, combinations, etc.)

Hi everyone, I'd like some help please

1) A basket contains 7 red apples and 4 yellow apples. A sample of 3 apples is drawn. Find the probabilities that the sample contains more red than yellow apples.

So, I figure I have to find the sum of the probabilities that the sample has:
-4 red apples and 3 yellow apples, 5 red apples and 2 yellow apples, 6 red apples and 1 yellow apple, and 7 and 0 right....right?

I want to use combinations, right? Because I know that when it asked for all red apples I did (4 3) divided by (11 3)

*pretend that those numbers are on top of each other like in combination form*



and my second question...
2) given a certain number of balls, of which some are blue, pick 5 at random. thie probability that all 5 are blue is 1/2. determine the original number of balls and decide how many were blue

This one, I have no clue.

Thanks!

For #1

Hi Plato,

I've been looking and trying to understand what you've written for the last hour...and i'm still not understanding. are you sure that that answer is correct? that would give me 119/165, or 72%. i thought i would add the (7 2) and (4 1), not multiply them!

Thanks

In order for a sample of three to contain more reds than yellows, then it contains all reds
or it contains two reds and one yellow .

Thank you so much!!

While I'm at it do you or anyone know how to solve the second problem? I've seen tutors, etc and nobody knows! Thanks!!

Originally Posted by gobbajeezalus

[snip]
2) given a certain number of balls, of which some are blue, pick 5 at random. thie probability that all 5 are blue is 1/2. determine the original number of balls and decide how many were blue

This one, I have no clue.

Thanks!

Let number of blue ball be x and total number of balls be n. Then .

Your job is to solve this equation under the conditions (i) x < n and (ii) x and n are positive whole numbers. I found a solution using trial and error in about 27 seconds (using a calculator).