the probability that the second one does not match the first is 5/6, the probability that the third does not match either one (assuming they all don't match) is 4/6, so the probability that all three don't match is 5/6*4/6=5/9, so the probability of a match is 1-5/9 or 4/9
Now, we have to take into account the possibility that the second and third match the first, which is 1/36, so the probability of exactly one match is 4/9-1/36=11/36
Thinking about this more...scratch the paragraph directly above. There are only three types of outcomes: there are no matches (prob of that is 5/9 or 20/36), prob of two and only two matching, and prob of all three matching. The prob of the second two choices matching the first is 1/36, leaving the prob of exactly one match as 1-20/36-1/36=15/36=5/12 as others have found.
You do realize that kuiper had the wrong answer until he read the right answers below and revised his?
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What is the probability that 2 M%26amp;Ms have the same color when 3 are drawn?Probability of your picking up one M%26amp;M of a particular color is (1/6).
Probability that you may pick up the second M%26amp;M of the same color = (1/6)
Probability that you may pick up the third M%26amp;M of a different color = (5/6)
Hence, probability that you may pick up first 2 M%26amp;Ms of a particular color and the third of a different color
= (1/6)^2 * (5/6) = 5/216
This can happen in 3 different ways
=%26gt; probability = 15/216 = 5/72
Hence, for 6 colors, probability that any 2 M%26amp;Ms are of the same color and the third of the different color
= 6 * (5/72)
= 5/12
I just got done taking statistics (statistics for engineers and scientists) at the university I go to, so I am pretty confident of that William's answer is right. His explanation is correct too, but I will explain in more mathematical terms.
We will call success P and a failure Q. P represents a that a particular color is chosen and probability is 1/6. Q is the probability that any other color is chosen and the prob is 5/6.
We will call the number of trials n and the number of successes x. You want to take three trials and achieve two successes.
According the binomial distribution, the probability of choosing x successes in n trials with probability of success P and failure Q is given by the following:
P(X=x)=(n choose x)(p^x)(q^x)
The choose term is a probability term that give the number of combinations of the answers. 3 choose 2 is three so the resulting probability is:
3(1/6)^2(5/6)=3(1/6)(1/6)(5/6)=6.9444%
By the way, no answer above 16.66% (1/6) makes any sense because that is probability that you would choose any one color.What is the probability that 2 M%26amp;Ms have the same color when 3 are drawn?
What we are attempting to get is:
2 with the same color, 1 of a different color.
First there are 216 combinations (6^3)
Of these we can form 2 the same and 1 different as follows:
AAB
ABA
BAA
We have 6 choices for the A color, 5 choices for the B color and 3 ways to have drawn them.
6 x 5 x 3 = 90
90 good outcomes
216 possible outcomes
90 / 216
= 10 / 24
= 5 / 12
To prove that this is the right answer, here's a full enumeration of the 90 ways to have picked 3 M%26amp;Ms where exactly 2 match color.
Red Red Orange
Red Red Yellow
Red Red Green
Red Red Blue
Red Red Brown
Red Orange Red
Red Orange Orange
Red Yellow Red
Red Yellow Yellow
Red Green Red
Red Green Green
Red Blue Red
Red Blue Blue
Red Brown Red
Red Brown Brown
Orange Red Red
Orange Red Orange
Orange Orange Red
Orange Orange Yellow
Orange Orange Green
Orange Orange Blue
Orange Orange Brown
Orange Yellow Orange
Orange Yellow Yellow
Orange Green Orange
Orange Green Green
Orange Blue Orange
Orange Blue Blue
Orange Brown Orange
Orange Brown Brown
Yellow Red Red
Yellow Red Yellow
Yellow Orange Orange
Yellow Orange Yellow
Yellow Yellow Red
Yellow Yellow Orange
Yellow Yellow Green
Yellow Yellow Blue
Yellow Yellow Brown
Yellow Green Yellow
Yellow Green Green
Yellow Blue Yellow
Yellow Blue Blue
Yellow Brown Yellow
Yellow Brown Brown
Green Red Red
Green Red Green
Green Orange Orange
Green Orange Green
Green Yellow Yellow
Green Yellow Green
Green Green Red
Green Green Orange
Green Green Yellow
Green Green Blue
Green Green Brown
Green Blue Green
Green Blue Blue
Green Brown Green
Green Brown Brown
Blue Red Red
Blue Red Blue
Blue Orange Orange
Blue Orange Blue
Blue Yellow Yellow
Blue Yellow Blue
Blue Green Green
Blue Green Blue
Blue Blue Red
Blue Blue Orange
Blue Blue Yellow
Blue Blue Green
Blue Blue Brown
Blue Brown Blue
Blue Brown Brown
Brown Red Red
Brown Red Brown
Brown Orange Orange
Brown Orange Brown
Brown Yellow Yellow
Brown Yellow Brown
Brown Green Green
Brown Green Brown
Brown Blue Blue
Brown Blue Brown
Brown Brown Red
Brown Brown Orange
Brown Brown Yellow
Brown Brown Green
Brown Brown Blue
P.S. To those that are getting 5/72 as an answer, you are forgetting that there is no designated color. It didn't say that you have to get 2 *red* and 1 non-red, for example. It said that exactly two have to be the same color. Multiply your answer by 6 (any of 6 colors as the double) and you get the correct answer of 30/72 = 5/12
the probablity is
3(1/6)(1/6)(5/6)
=15/216
to get a colour,you have 1/6 probablity.
to get the same colour for the second M%26amp;M,you have 1/6 probablity.
to get a different colour for the third M%26amp;M,you have 1-1/6=5/6 probablity.
the events are consecuitve,hence the values are multiplied.This can occur 3 times,that is either first,second and third is not the same with others.Hence,the value is multiplied by 3.
OK assuming that the bag has the same number of each color your chance of pulling 2 of the same color and one of a different color would be 5 out of 12
ok. your first M%26amp;M can be any color, so the probability of picking it is just 1.
Now, there are two cases for your second one.
You pick the color, or you don't.
Ok, so if you do- there is 1/6 chance that you pick it, and then a 5/6 chance that you don't pick it the second time, which is 5/36.
If you don't, there is a 5/6 chance of that happening, and then a 1/3 chance of getting the right one, which is 5/18=10/36.
5/36+10/36=15/36=5/12
The first M%26amp;M can be any color.
There are two cases for the second M%26amp;M drawn.
p(different color than first) = 5/6
p(same color as first) = 1/6
____________
If the first two M%26amp;Ms are different colors the chance of matching one of them with the third M%26amp;M is
p(match) = 2/6 = 1/3
If the two M%26amp;Ms are the same color the chance of drawing a different colof with the third M%26amp;M is
p(no match) = 5/6
__________
Put this together and you get the probability of two and only two colors between three M%26amp;Ms
P(2 colors) = (5/6)(1/3) + (1/6)(5/6) = 5/18 + 5/36 = 5/12
Assuming there are only six colors and there is equal probability for each color then
Let X be the number of M%26amp;M of a given color. X has the binomial distribution
with n = 3 trials and success probability p = 0.1666667 .
In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x.
this is found by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.
Or, to be more accurate, the binomial is the sum of n independent and identically distributed Bernoulli trials.
X ~ Binomial( n , p )
the mean of the binomial distribution is n * p = 0.5
the variance of the binomial distribution is n * p * (1 - p) = 0.4166667
the standard deviation is the square root of the variance = 0.6454972
The Probability Mass Function, PDF,
f(X) = P(X = x) is:
P(X = 0 ) = 0.578703703703704
P(X = 1 ) = 0.3472222222222222
P(X = 2 ) = 0.06944444444444445
P(X = 3 ) = 0.00462962962962963
Now that is just for one color, we have to consider the possibility of any color appearing twice.
There are six possible colors and so we multiply:
6 * P(X = 2) = 0.4166667
to confirm this calculation I wrote a simulation in R to draw three MM's from an infinite bag with equal probability of getting any one color. The code counts the number of times that the there are two MM's of the same color selected
MM %26lt;- seq(1,6,1)
count %26lt;- 0
trials %26lt;- 5500
for(i in 1:trials)
{
data %26lt;- sample(MM,3,TRUE)
mode %26lt;- sort(-table(data))
if(mode[[1]] == -2) {count %26lt;- count + 1}
}
cat("counts:",count,"\nProb:",count/tr鈥?br>
OUTPUT:
counts: 2287
Prob: 0.4158182
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