Question

Two fair dice are thrown. Let the random variable X be the smaller of the two scores if the scores are different, or the score on one of the dice if the scores are the same
The full question is here:
http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s04_qp_6.pdf
Question No. 3
Explaination and workings please! :)

Answer 1

I think you should consider all the scores for when "1" is the smallest number (or equal to the 2nd die's number, i.e. "1") and count all those possibilities:
(1, 1), (1, 2), ... (1, 6), (2, 1), (3, 1), .., (6, 1) which should be 11 possibilities. And since the roll of 2 dice has a sample space of 36 outcomes, then P(X=1) = 11/36

You can do the same for the number "2" "
(2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (4, 2), (5, 2), (6, 2), so 9 possibilities, meaning P(X=2) = 9/36

The pattern should continue in this way where there are two-less possibilities for every +1 number you consider on the die.

Then, the expectation is:
\[\large E(X)=\sum_{x=1}^6x P(X=x)=1\left(\frac{11}{36} \right)+2\left(\frac{9}{36} \right) +...\text{sum until 6th term} \]

Answer 2

I don't really get it for p(x=2) why can't be (2,1) part of it?

Answer 3

You want to consider all possibilities where "2" is the smallest number (or equal number) of the two. So here "1" is smaller than 2, so we can't consider that possibility since we want the ones where "2" is the smallest

Answer 4

I see. Thank so much! I got it now. :)

Answer 5

yay :)