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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!

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    $\begingroup$ look up inclusion exclusion principle $\endgroup$ – Daniel Xiang Jan 18 '17 at 18:01
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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.

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

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