Answer to Question #323698 in Statistics and Probability for Klum

Question

Two dice are thrown. What is the probability that the first die rolls a 1 if it is known that the number of points on the second die is higher than on the first?

Answer 1

Let's A - the event of rolling a 1 of the first die.

Obviously the number of points on the second die can take on one of the values 2, 3, 4, 5, 6.

For convenience let's denote the respective events B₂, B₃, B₄, B₅, B₆.

These events are equally possible, $P(B_2)=P(B_3)=P(B_4)=P(B_5)=P(B_6)=\cfrac{1}{5}.$

According to the law of Total Probability,

$P(A)=P(B_2)\cdot P(A|B_2)+P(B_3)\cdot P(A|B_3)+\\ +P(B_4)\cdot P(A|B_4)+P(B_5)\cdot P(A|B_5)+\\ +P(B_6)\cdot P(A|B_6).$

Here $P(A|B_i)$ are the conditional probabilities.

If there are 2 points on the second die, $P(A|B_2)=1,$ because the only event when the number of points on the first die is less then 2 is A.

For the event B₃ there are possible values for the number of points on the first die are 1 and 2,

$P(A|B_3)=\cfrac{1}{2}.$

For the event B₄ there are possible values for the number of points on the first die are 1, 2 and 3,

$P(A|B_4)=\cfrac{1}{3},$ and so on.

The sought probability of 1 point on the first die:

$P(A)=\cfrac{1}{5}\cdot 1+\cfrac{1}{5}\cdot \cfrac{1}{2}+\cfrac{1}{5}\cdot \cfrac{1}{3}+\cfrac{1}{5}\cdot \cfrac{1}{4}+\cfrac{1}{5}\cdot \cfrac{1}{5}=\\ =\cfrac{1}{5}\cdot \cfrac{137}{60}= \cfrac{137}{300}\approx0.457.$