Probability with two different variables There are 11 yellow balls and 16 green balls in a box. Jacob randomly pulls two balls, one at a time. What is the probability that Jacob will have pulled two different colored balls?
Here's what I have:
There are (11 * 10) + (11 * 16) + (16 * 15) = 526 total outcomes
11 * 16 = 176 outcomes of two different colored balls
So the probability is 176/526 = 0.33 or 33%
Did I get this question right?
 A: Not quite. You can get the correct answer by using an alternative method.
Ordered samples: This method keeps track of the order in which the balls are withdrawn from the box:
$$P(\text{Diff Col}) = P(YG) + P(GY),$$ and $P(YG) = (11/27)(16/26).$
This is an application of the 'general multiplication rule':
$P(AB) = P(A)P(B|A).$ That's a good start. I'll let you finish it to get $352/702.$
Unordered samples: To fix your method, I think you want $\frac{{11\choose 1}{16\choose 1}}{{27\choose 2}} = 176/351 = 0.5014.$ [Note: This method does not keep track of
the order in which the balls are drawn--only how many of them are of
each color.]
Hypergeometric distribution: This can also be solved using a hypergeometric random variable $X$ that
counts the yellow balls when two balls are drawn from the box you describe.
In R statistical software the PDF of this distribution is denoted 'dhyper'.
dhyper(1, 11, 16, 2)
## 0.5014245

Maybe you will cover hypergeometric distributions later in your course. The support of $X$ is $\{0, 1, 2\}$ and the PDF table can be
found as follows:
x = 0:2;  pdf=dhyper(x, 11, 16, 2)
cbind(x, pdf)
##  x       pdf
##  0 0.3418803
##  1 0.5014245
##  2 0.1566952


Simulation: Finally just for fun, here is a simulation of a million such two-ball
experiments in R. Yellow balls are 0's and green balls
are 1's; the sum is 1 precisely when balls of different colors are drawn.
Simulation is not exact, but here it gives three-place accuracy.
box = c(rep(0,11), rep(1,16))
x = replicate( 10^6, sum(sample(box, 2)) )
mean(x==1)
## 0.501186

A: You would have $11+16=27$ choose $2$ which is $351$, and your calculation of $11\cdot 16$ is correct, so you would have $\frac{176}{351}=0.50142$
