Show that P(A1 ∪ A2 ∪ A3 ∪ A4) = 1 − (1 − (1/4))^4

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;

$P(A_i ∩ A_j) = (1/4)^2$ , for i 6= j;

$P(A_i ∩ A_j ∩ A_k) = (1/4)^3$ , for i, j, k all different; and

$P(A_1 ∩ A_2 ∩ A_3 ∩ A_4) = (1/4)^4$

Show that $P(A_1 ∩ A_2 ∩ A_3 ∩ A_4) = 1-(1-(1/4))^4$

for the above question do i just solve the right side? or is there any other way to show $P(A_1 ∩ A_2 ∩ A_3 ∩ A_4) = 1-(1-(1/4))^4$?

Thank you!

• look up inclusion exclusion principle – Daniel Xiang Jan 18 '17 at 18:01

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}
• How did you get $1-(1-(1/4))^4$ – ISuckAtMathPleaseHELPME Jan 18 '17 at 22:00
• $P(\overline{Ai}) = 1-P(Ai) = 1-\frac{1}{4}$ for each $i$, and because of independence of dice rolls, we multiply the probabilities to get the probability of the intersection – Blaza Jan 18 '17 at 22:16