Probability that X = 1 where X is minimum of two random variables when a dice is roled.

Question

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. Then, what is the probability that X = 1 and Also, what is E(X)

I thought , since its minimum of X1 and X2, it can be either X1 is 1 or X2 is 1, so, answer is 2/36. but, looks like i am wrong. please help

Answer 1

$$P(X=1)=P(X_1=1\vee X_2=1)=P(X_1=1)+P(X_2=1)-P(X_1=1\cap X_2=1)=$$$$P(X_1=1)+P(X_2=1)-P(X_1=1)P(X_2=1)=\frac16+\frac16-\frac16\frac16$$

Or:

$$P(X=1)=1-P(X>1)=1-P(X_1>1\wedge X_2>1)=$$$$1-P(X_1>1)P(X_2>1)=1-\frac56\frac56$$

---

Hint for expectation.

If $X$ is a random variable taking values in $\{0,1,2,\dots\}$ then:$$\mathbf EX=\sum_{k=1}^{\infty}P(X\geq k)$$