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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.

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  • $\begingroup$ Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Sep 17 '19 at 8:34
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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 $4$: $\{(1,5),(2,4),(3,3),(4,2),(4,6),(5,1),(5,5),(6,4)\}$. $\lvert S\cap T\rvert=10$

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  • $\begingroup$ Thank you for your help. $\endgroup$ – Melissa E Sep 18 '19 at 4:15
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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. $P(S)P(T)=\frac 12 \cdot \frac {16}{36}=\frac 8{36}$

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

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