1. ## Another Small Problem

Hello there, I have met another small problem that I need some help

Suppose you have an 40-card deck. 3 of the cards are marked. You draw 6 cards. What are the chances of having 1 of these marked cards in your 6-card hand?

Thanks so much

2. Originally Posted by Kostas
Hello there, I have met another small problem that I need some help

Suppose you have an 40-card deck. 3 of the cards are marked. You draw 6 cards. What are the chances of having 1 of these marked cards in your 6-card hand?

Thanks so much
This is an example of the hypergeometric distribution. But it can also be done using an understanding of combinatorics. The answer is $\displaystyle \frac{^3C_1 \cdot ^{37}C_5}{^{40}C_6}$.

Study this answer. Do you see where the different bits of it have come from?

3. Ok this gives around 34%.

I'd think something like that: 3/40*39/39*38/38*37/37*36/36*35/35 which gives around 7.5% which I think it's wrong Can the solution be written in somewhat this way? (and generally how is this way called so I can read it somewhere?)

4. Hello, Kostas!

Your fractions are off . . .

There are 3 Marked cards and 37 Unmarked cards.

Suppose we want the probability of $\displaystyle MUUUUU$ in that order.

The probability is: .$\displaystyle \frac{3}{40}\cdot\frac{37}{39}\cdot\frac{36}{38}\c dot\frac{35}{37}\cdot\frac{34}{36}\cdot\frac{33}{3 5}\;=\;\frac{561}{9880}$

Since $\displaystyle M$ can appear in any of the six positions:

. . $\displaystyle P(\text{one M}) \;=\;6\times \frac{561}{9880} \;=\;\frac{1683}{4940} \;\approx\; 34\%$

5. Cool that was exactly what I needed
Ok I've understood everything now and the problem is solved! Thanks so much!
Just tell me how is this second method called (with the P(one M) thing) so I can find more about it!