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!
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!
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