- #1
Gridvvk
- 56
- 1
Homework Statement
You toss a fair die 4 times. You let the random variable X denote the number of times a toss results in 1 or 2, and Y denote the number of time a toss results in 3,4, or 5.
a) Find P(X = 1, Y = 2)
b) Find P(X = Y)
Homework Equations
None that I know of.
The Attempt at a Solution
I defined a third r.v. Z := number of times a toss results in a 6.
X ~ Bin(4,1/3), Y~Bin(4,1/2), Z~Bin(4,1/6).
a) P(X = 1, Y = 2) = P(X = 1, Y=2, Z = 1) = (4 choose 1)(3 choose 1)(1/3)(1/2)^2(1/6) = 1/6
You needed to take the groupings into account. There are 12 such orderings, one being
XYYZ, my reasoning was there are (4 choose 1) ways to move that Z around, and (3 choose 1 ways) to move X in XYY, giving 4 * 3 = 12 different possibilities.
b) P(X = Y).
Case 1: X=Y=0, Z = 4
P(Case 1) = (1/6)^4
Case 2: X=Y=1, Z = 2
P(Case 2) = 2(4 choose 2)(1/3)(1/2)(1/6)^2 = 1/18
Where you have 12 orderings again here, one such is XYZZ, two ways to shift the XY, and (4 choose 2) ways to work with the ZZ.
Case 3: X = Y = 2, Z = 0.
P(Case 3) = (3 choose 1)(2 choose 1)(1/3)^2(1/2)^2 = 1/6
Where you have 6 groupings, one such is XXYY, my reasoning was you can move the second X in XYY (3 choose 1) ways and the outer X would then have (2 choose 1) ways.
So I got P(X = Y) = P(Case 1) + P(Case 2) + P(Case 3) = 289 / 1296.
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I think my answer is correct (I could be wrong though). However, is there a better way to do this problem using our knowledge that X,Y, and Z are binomial? I only ask because I was a bit uncertain on the number of arrangements and just did it solely on intuition and oftentimes enumerated all possibilities to see if I was correct, so is there a better way to count the number of possibilities as well?