MHB Probability of drawing mutiple marbles.

AI Thread Summary
The discussion revolves around calculating the probability of drawing at least one red and one blue marble from a bag containing 60 marbles of different colors. The user seeks a formula to apply various values for red, blue, and yellow marbles, specifically using the numbers X=4, Y=12, and Z=44. A suggested approach involves using the complement probability method, where the probability of drawing no red or no blue marbles is calculated first. By applying the inclusion/exclusion principle, the user can find the desired probability by subtracting the complement from one. The conversation emphasizes the importance of understanding "at least" in probability calculations.
billy1
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Hi, first post here...

Wasn't sure to put this in Basic Probability and Statistics or Advanced Probability and Statistics, figured I'd put it here in Advanced so it can get better help, if this belongs in Basic, mods feel free to move it.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!
 
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billy1 said:
Wasn't sure to put this in Basic Probability and Statistics or Advanced Probability and Statistics, figured I'd put it here in Advanced so it can get better help, if this belongs in Basic, mods feel free to move it.

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.
What is the probability of drawing no red marbles or drawing no blue marbles.
Use inclusion/exclusion. Then find the complement.
 
Hi billy1,

Welcome to MHB! :)

Whenever you see the words "at least" in probability a helpful way of calculating this is to realize that "at least 1 of something" is "1 - probability of none". Put more mathematically, [math]P[X>0]=1-P[X=0][/math]. Since this is discrete probability, $P[X>0]$ is the same thing as saying $P[X \ge 1]$. It doesn't matter which way you write it.

So as Plato wrote, you can do this problem by looking at "1-complement", or the opposite probability, for each event you wish to happen.

Jameson
 
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