# Combinations Probability Problem

• January 13th 2011, 12:04 PM
mathguy20
Combinations Probability Problem
Any help on how to do the following probability problem would be appreciated.

Thanks

Quote:

I have planted new tulip bulbs in my front yard, but confused the colors of the bulbs. If I had 8 yellow, 6 red, 5 pink, and 5 purple tulips, what is the probability that 6 bulbs I planted will all be the same color?
• January 13th 2011, 12:16 PM
Quote:

Originally Posted by mathguy20
Any help on how to do the following probability problem would be appreciated.

Thanks

Clearly they will need to be all yellow or all red.

Calculate the probability of either event happening independently.

The easier one is to calculate the probability that all 6 will be red.

Can you evaluate that?
• January 13th 2011, 12:21 PM
Pranas
Quote:

Originally Posted by mathguy20
Any help on how to do the following probability problem would be appreciated.

Thanks

I have planted new tulip bulbs in my front yard, but confused the colors of the bulbs. If I had 8 yellow, 6 red, 5 pink, and 5 purple tulips, what is the probability that 6 bulbs I planted will all be the same color?

I see it like this:

All 6 yellow: $$\mathop C\nolimits_8^6 = \frac{{8!}}{{6! \cdot (8 - 6)!}} = 28$$
All 6 red: $$\mathop C\nolimits_6^6 = 1$$
Successful combinations: $$m = 28 + 1 = 29$$

Total combinations (successful + unsuccessful): $$n = \mathop C\nolimits_{24}^6 = \frac{{24!}}{{6! \cdot (24 - 6)!}} = 134596$$

Therefore probability $$P = \frac{m}{n} = \frac{{29}}{{134596}}$$
• January 13th 2011, 12:34 PM
mathguy20
Quote:

Clearly they will need to be all yellow or all red.

Calculate the probability of either event happening independently.

The easier one is to calculate the probability that all 6 will be red.

Can you evaluate that?

Is it 6/24 that all are red and 8/24 that all are yellow?
• January 13th 2011, 12:41 PM
Quote:

Originally Posted by mathguy20
Is it 6/24 that all are red and 8/24 that all are yellow?

There are 24 bulbs in total, so that's the denominator for the probabilities.
You are correct for the red,
but you need Pranas' logic for the yellow.

If you plant 6 bulbs and they are all yellow,
it could be some set of 6 from the 8 yellow.

So you must select 6 of 8 first to calculate the probability of planting 6 yellow.
How many ways are there to choose 6 from 8 ?