Answer
$P(E or F or G)=P(E)+P(F)+P(G)-P(E and F)-P(E and G)-P(F and G)+P(E and F and G)$
Work Step by Step
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24}. So, N(S) = 24
E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 14}. So, N(E)=12
F = {5, 6, 7, 8, 9, 10, 12, 13, 15, 16}. So, N(F)=10
G = {8, 9, 10, 11, 12, 13, 14, 17, 18, 19, 20}. So, N(G)=11
E or F or G = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}. So, N(E or F or G)=20
E and F = {5, 6, 7, 8, 9, 10}. So N(E and F) = 6
E and G = {8, 9, 10, 11, 14}. So, N(E and G) = 5
F and G = {8, 9, 10, 12, 13}. So, N(F and G) = 5
E and F and G = {8, 9, 10}. So, N(E and F and G) = 3
In the sum $N(E) + N(F) + N(G) = 33 \ne N(E or F or G)$ beacuse we counted the outcomes 5, 6, 7, 11, 12, 13, 14 twice and 8, 9, 10 three times.
Now, $N(E) + N(F) + N(G) - N(E and F) - N(E and G) - N(F and G) = 17 \ne N(E or F or G)$
because the outcomes 8, 9, 10 have been removed 3 times, but they should have been removed only twice.
To correct the answer: $N(E) + N(F) + N(G) - N(E and F) - N(E and G) - N(F and G) + N(E and F and G) = 20 = N(E or F or G).$
Now, we have:
$N(E or F or G) = N(E) + N(F) + N(G) - N(E and F) - N(E and G) - N(F and G) + N(E and F and G).$
Dividing both sides by N(S):
$\frac{N(E or F or G)}{N(S)}=\frac{N(E)}{N(S)}+\frac{N(F)}{N(S)}+\frac{N(G)}{N(S)}-\frac{N(E and F)}{N(S)}-\frac{N(E and G)}{N(S)}-\frac{N(F and G)}{N(S)}+\frac{N(E and F and G)}{N(S)}.$
But, $\frac{N(E)}{N(S)}=P(E)$, $\frac{N(F)}{N(S)}=P(F)$, etc. So:
$P(E or F or G)=P(E)+P(F)+P(G)-P(E and F)-P(E and G)-P(F and G)+P(E and F and G).$
