# Better way to do this probability problem?

1. Nov 9, 2013

### Gridvvk

1. The problem statement, all variables and given/known data
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)

2. Relevant equations
None that I know of.

3. 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?

2. Nov 9, 2013

### Ray Vickson

Both of your answers are correct. You could have saved a bit of time by using the trinomial distribution: if each 'trial' results in either outcomes A, B or C with respective probabilities $p_A,\: p_B, \: p_C \; (p_A + p_B + p_C = 1)$, then for n independent trials, we have
$$P\{ N_A = a, N_B = b, N_C = c\} = \frac{n!}{a! \: b! \: c!}\, p_A^a \, p_B^b \, p_C^c$$
where $a+b+c=n$.

You can easily derive this by successive application of the binomial:
$$P\{N_A=a, N_B=b, N_C=c \} = P\{N_A=a\} P\{N_B=b, N_C=c | N_A = a\}.$$
We have
$$P\{N_A=a\} = {n \choose a} p_A^a \; (p_B + p_C)^{n-a},$$
and
$$P\{N_B=b, N_C=c | N_A = a\} = {n-a \choose b} q_B^b \, q_C^c,$$
where
$$q_B = P\{B| B \text{ or }C\} = \frac{p_B}{p_B + p_C} \\ q_c = P\{C| B \text{ or }C\} = \frac{p_C}{p_B + p_C}.$$
If you put all these together you get the trinomial formula.

3. Nov 9, 2013

### LCKurtz

I haven't worked your problem, but you can use the trinomial distribution:$$P(X=x,Y=y) = \frac {4!}{x!y!(4-x-y)!}\left(\frac 1 3\right)^x\left(\frac 1 2\right)^y \left(1-\frac 1 3 -\frac 1 2\right)^{(4-x-y)}$$

: I see Ray beat me to it by a full minute.

Last edited: Nov 9, 2013
4. Nov 9, 2013

### Gridvvk

Thanks! That does make the calculations easier and helps me on another similar problem.