# Thread: Another Small Problem

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?

It seems so easy but I am stuck here for hours please help me!
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?

It seems so easy but I am stuck here for hours please help me!
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 $\frac{^3C_1 \cdot ^{37}C_5}{^{40}C_6}$.

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 $MUUUUU$ in that order.

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

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

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

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!