Advanced Probability Question.

billy1

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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!
 
Last edited:
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.
 
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 \(\displaystyle 1-\frac{\binom{Y+Z}{7}}{\binom{X+Y+Z}{7}}-\frac{\binom{X+Z}{7}}{\binom{X+Y+Z}{7}}+
\frac{\binom{Z}{7}}{\binom{X+Y+Z}{7}}\)
 
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