# Two ordinary fair dice are thrown and the numbers obtained are noted. Determine whether events $S$ and $T$ are independent?

Two ordinary fair dice are thrown and the numbers obtained are noted. Event S is ‘The sum of the numbers even’. Event T is ‘The sum of the numbers is either less than 6 or a multiple of 4 or both’. Determine whether S and T are independent?

So I know that:

$$P(S)= 1/2$$

$$P(T)= 16/36$$

And for independent events we use: $$P(S \cap T) = P(S)P(T)$$

When I check my answer with the mark scheme it’s incorrect and the answer in the MS for $$P(S \cap T)$$ is $$10/36$$.

I’m not sure which part I’m doing wrong...I would be grateful if you could give a detailed answer for calculating the independent events part.

• Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Sep 17 '19 at 8:34

For the sum to be even, there are three favoured results of the second die for each result of the first die. $$\lvert S\rvert = 18$$

From this we exclude the eight outcomes where the sum is also not less than six, and not a multiple of $$4$$: $$\{(1,5),(2,4),(3,3),(4,2),(4,6),(5,1),(5,5),(6,4)\}$$. $$\lvert S\cap T\rvert=10$$

• Thank you for your help. – Melissa E Sep 18 '19 at 4:15

The events in $$S \cap T$$ are sums of $$2,4,8,12$$, with probabilities of $$\frac 1{36}, \frac {3}{36}, \frac 5{36}, \frac 1{36}$$. These sum to $$\frac {10}{36}$$ as the answer sheet says. $$P(S)P(T)=\frac 12 \cdot \frac {16}{36}=\frac 8{36}$$

• Thank you for your help. – Melissa E Sep 18 '19 at 4:15