A pair of fair dice is tossed. Find the probability that the maximum of the two numbers is greater than $4$.
My attempted solution:
$E=\{(1,5), (2,5), (3,5), (4,5), (5,5), (6,5),$
$(1,6), (2,6), (3,6), (4,6), (5,6), (6,6),$
$(5,1), (5,2), (5,3), (5,4), (5,6),$
$(6,1), (6,2), (6,3), (6,4), (6,5)\}$
Here $n_s = 22.$
So, $P =\frac{22}{36} = \frac{11}{18}.$
But, the correct answer is, $\frac{5}{9}$.
What am I missing?
For it to be false, both dies have to be $1,2,3,4$. $$ p = \frac{4}{6} \times \frac{4}{6} = \frac{16}{36} = \frac{4}{9}, $$ therefore the probability of the statement being true is $$ 1 - \frac{4}{9} = \frac{5}{9}. $$
To calculate without using the complement, if the dies are $X$ and $Y$, it is $$ P(X > 4) + P(Y > 4) - P(X>4,\ Y>4) = \frac{1}{3} + \frac{1}{3} - \frac{1}{9} = \frac{6}{9} - \frac{1}{9} = \frac{5}{9}. $$ It is slightly trickier, if you add the two probabilities together, you have to knock off the area where they are both true. You can draw a Venn diagram to see that $$\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$ is the probability that both dice show greater than $4$.