Roll a fair four-sided die four times. Let $A_i$ be the event that side i is observed on
the ith roll: this is referred to as a match on the ith roll. It is given that $P(A_i) = 1/4$
for each i = 1, 2, 3, 4;
$P(A_i \cap A_j) = (1/4)^2$
, for $i \neq j$;
$P(A_i \cap A_j \cap A_k) = (1/4)^3$
, for
i, j, k all different; and
$P(A_1 \cap A_2 \cap A_3 \cap A_4) = (1/4)^4$
Show that $P(A_1 \cup A_2 \cup A_3 \cup A_4) = 1-(1-(1/4))^4$
for the above question do i just solve the right side? or is there any other way to show $P(A_1 \cup A_2 \cup A_3 \cup A_4) = 1-(1-(1/4))^4$?
Thank you!
Best Answer
Assuming that dice rolls are independent, to show that $P\left(A_1\cup A_2\cup A_3 \cup A_4\right)=1-(1-(1/4))^4$ you can just use De Morgan's laws and write $\overline{A_1\cup A_2\cup A_3 \cup A_4} = \overline{A_1}\cap\overline{A_2}\cap\overline{A_3}\cap\overline{A_4}$ and so the probability is \begin{align} P\left(A_1\cup A_2\cup A_3 \cup A_4\right)&=1-P\left(\overline{A_1\cup A_2\cup A_3 \cup A_4}\right)\\ &=1 - P\left(\overline{A_1}\cap\overline{A_2}\cap\overline{A_3}\cap\overline{A_4}\right)\\ &= 1- \left(1-\left(\frac{1}{4}\right)\right)^4 \end{align}
Edit: This is one possible solution, but as Daniel Xiang implied in the comments, the problem is formulated to encourage the use of inclusion–exclusion principle, so maybe that's the better way to go.