Advanced Probability Question.

[billy1]

There are 60 Marbles in a bag. There are X red marbles, Y blue marbles and Z yellow marbles.

X+Y+Z=60

If 7 Marbles are drawn from the bag (without replacement) what is the probability that at least 1 marble is red and 1 marble is blue. The order of drawing does not matter. The other 5 marbles can be red, blue or yellow, we only want to find out if at least 1 of the 7 marbles is red and 1 of the 7 marbles is blue.

My goal is to find a formula I can use to plug in different numbers for X, Y and Z. This is not for school, this is for a trading card game I play, I changed it to Marbles so you won’t get confused as the game does not use a standard 52 card deck, it uses 60 cards (hence marbles). Not sure if anyone here is familiar with trading card games.

The numbers I have been playing around with is X=4, Y=12, and Z=44.

This is what I did so far-

(56/60)*(55/59)*(54/58)*(53/57)*(52/56)*(51/55)*(50/54)

Above should be the 7 draws of the deck where a red marble (X=4) is not drawn.

Subtract the result I get from above from 1 gives me the probability of drawing 1 red marble off of 7 draws. Below is the formula for.

(1-(((Y+Z)-0)/60)*( (Y+Z)-1)/59)*( (Y+Z)-2)/58)*( (Y+Z)-3)/57)*( (Y+Z)-4)/56)*( (Y+Z)-5)/55)*( (Y+Z)-6)/54)))

Not even sure if this is even in the right direction to get what I am looking for, please help!

 

[daon2]

There are 60 Marbles in a bag. There are X red marbles, Y blue marbles and Z yellow marbles.

X+Y+Z=60

If 7 Marbles are drawn from the bag (without replacement) what is the probability that at least 1 marble is red and 1 marble is blue. The order of drawing does not matter. The other 5 marbles can be red, blue or yellow, we only want to find out if at least 1 of the 7 marbles is red and 1 of the 7 marbles is blue.

My goal is to find a formula I can use to plug in different numbers for X, Y and Z. This is not for school, this is for a trading card game I play, I changed it to Marbles so you won’t get confused as the game does not use a standard 52 card deck, it uses 60 cards (hence marbles). Not sure if anyone here is familiar with trading card games.

The numbers I have been playing around with is X=4, Y=12, and Z=44.

This is what I did so far-

(56/60)*(55/59)*(54/58)*(53/57)*(52/56)*(51/55)*(50/54)

Above should be the 7 draws of the deck where a red marble (X=4) is not drawn.

Subtract the result I get from above from 1 gives me the probability of drawing 1 red marble off of 7 draws. Below is the formula for.

(1-(((Y+Z)-0)/60)*( (Y+Z)-1)/59)*( (Y+Z)-2)/58)*( (Y+Z)-3)/57)*( (Y+Z)-4)/56)*( (Y+Z)-5)/55)*( (Y+Z)-6)/54)))

Not even sure if this is even in the right direction to get what I am looking for, please help!

Magic? If so, I would err on the side of caution.

Anyway, you cannot calculate the exact number without knowing more. You can obtain an answer in terms of X,Y,Z though.

You want

$\displaystyle \displaystyle \dfrac{1}{{60\choose 7}}\sum_{i=1}^6\sum_{j=1}^{7-i} {X\choose i}{Y\choose j}{60-X-Y \choose 7-i-j}$

Where you may substitute Z=60-X-Y.

I am no expert in probability, so there may be a better way.

 

[pka]

There are 60 Marbles in a bag. There are X red marbles, Y blue marbles and Z yellow marbles. X+Y+Z=60

If 7 Marbles are drawn from the bag (without replacement) what is the probability that at least 1 marble is red and 1 marble is blue.

Say $\displaystyle X,~Y,~Z$ are all at least 7.

So \(\displaystyle 1-\frac{\binom{Y+Z}{7}}{\binom{X+Y+Z}{7}}-\frac{\binom{X+Z}{7}}{\binom{X+Y+Z}{7}}+
\frac{\binom{Z}{7}}{\binom{X+Y+Z}{7}}\)