An urn contains four fair dice. Two have faces numbered 1, 2, 3, 4, 5, and 6; one has faces numbered 2, 2, 4, 4, 6, and 6; and one has all six faces numbered 6. One of the dice is randomly selected from the urn and rolled. The same die is rolled a second time. Calculate the probability that a 6 is rolled both times.
My attempt
1/4[(1/6*1/6)+(1/6*1/6)+(2/6*2/6)+1]=7/24 1/4 Because one die would be chosen out of 4 Then if either of the regular die is chosen, 6 has prob of appearing 1/6 twice If it's the die with 2,2,4,4,6 and 6 then 6 has prob of appearing 2/6 twice Last the die with 6 all through has prob 1.
Let the events we pick one the regular dice be called $r$ )and the one where we pick the third $e$ (for even) and where we pick the final one $s$ (for sixes).
So $$P(r)=\frac12, P(e)=\frac14 = P(s)$$
Now given the die type $P(66)= p_6^2$ by independence where $p_6$ is the chance that one die lands on $6$.
So $$P(66|r)=\frac{1}{36}, P(66|e)=\frac{1}{9}, P(66|s)=1$$
Now apply the law of total probability:
$$P(66) = P(66|r)P(r) + P(66|e)P(e) + P(66|s)P(s)$$