Probability of drawing mutiple marbles.

In summary: Rule of 72 states that the probability of an event happening in a certain number of years is $1/72$ of that event happening. So, for example, if you want to know the probability of an event happening in the next 5 years, divide 1 by 72 to get the answer of 0.0625.In summary, the probability of drawing at least 1 red marble from 7 draws is 0.0625.
  • #1
billy1
1
0
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!
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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, \(\displaystyle P[X>0]=1-P[X=0]\). 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
 

What is the probability of drawing multiple marbles with replacement?

The probability of drawing multiple marbles with replacement is the same as the probability of drawing a single marble, since each draw is independent from the others. This means that the probability of drawing a specific marble remains constant with each draw.

What is the probability of drawing multiple marbles without replacement?

The probability of drawing multiple marbles without replacement changes with each draw, since the pool of remaining marbles decreases. This means that the probability of drawing a specific marble decreases with each draw.

How do you calculate the probability of drawing multiple marbles?

The probability of drawing multiple marbles is calculated by multiplying the individual probabilities of each draw. This means that if you want to find the probability of drawing a red marble and then a blue marble, you would multiply the probability of drawing a red marble by the probability of drawing a blue marble.

What is the difference between theoretical probability and experimental probability in regards to drawing multiple marbles?

Theoretical probability is based on mathematical calculations and assumes that all outcomes are equally likely. Experimental probability is based on actual data collected from trials and can vary depending on the sample size and conditions. In regards to drawing multiple marbles, the theoretical probability would be calculated using the total number of marbles and the desired outcome, while the experimental probability would be calculated using the results of actual draws.

How does the total number of marbles and the number of desired outcomes affect the probability of drawing multiple marbles?

The total number of marbles and the number of desired outcomes directly affect the probability of drawing multiple marbles. If there are more desired outcomes, the probability of drawing them decreases. If there are more marbles in total, the probability of drawing a specific marble decreases. This is because the larger the pool of marbles, the more possible outcomes there are and the less likely it is to draw a specific marble.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
9
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
12
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
790
  • Engineering and Comp Sci Homework Help
Replies
3
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
10
Views
1K
Replies
3
Views
2K
  • Precalculus Mathematics Homework Help
Replies
6
Views
2K
  • Precalculus Mathematics Homework Help
Replies
4
Views
798
Back
Top