[Math] A fair die is rolled twice. Let X1 and X2 denote the outcomes, and define random variable X to be the minimum of X1 and X2. What is Var(X)

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This question asks 4 parts. I can get them all except for the VAR. And I think I tried literally every combo and don't have it right. Here is where I'm at:

Q1: How many possible values are there for X?

A1: 6 (pretty easy, min can always be 1-6)

Q2: What is the probability that X=1

A2: 11/36 (This is the probability that at least one of the two rolls is a 1.)

Q3: What is E(X)?

Expected Probability is the Summation of the P(i) * i

for each number, find the combination of rolls that allow for your "i" to be the min. for example. A 5's possibilities are (5,5) – (5,6) and (6,5) – therefore 3 of 36 roll combinations would be valid for a 5 being lowest.

  • 1 * 11/36 = 11/36
  • 2 * 9/36 = 1/2
  • 3 * 7/36 = 7/12
  • 4 * 5/36 = 5/9
  • 5 * 3/26 = 5/12
  • 6 * 1/36 = 1/6

A3: (summation) = 91/36

So far so good 🙂

Q4: Var(X)

VAR = E(X-mu)^2

I've tried to compute this with the above E(X) in so man ways I'm spinning. Obviously, I'm close but I just can't get this. Can someone point me in the right direction to calculate the Var(X)?

Best Answer

For variance you know the formula. So find the $X-\mu_X$ for each X where $\mu_X $ is the mean or E(X) = 91/36. You can have columns like this: $X. |. f(X-\mu_X). $ where f is the frequency of each X (eg.instead of adding 1 eleven times , you can multiply by 11) So now find $E(X-\mu_X)^2$. Which is average. So simply add the entries in last column,square the number, and divide by $\sum f$ which is 36. You will get the answer

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