You have a shuffled deck of three cards: 2, 3, and 4, and you deal out the three cards. Let $E_{i}$ denote the event that ith card dealt is even numbered. (a) What is $\mathrm{P}\left[E_{2} | E_{1}\right]$, the probability the second card is even given that the first card is even? (b) What is the conditional probability that the first two cards are even given that the third card is even? Let $O_{i}$ represent the event that the ith card dealt is odd numbered. What is $\mathrm{P}\left[E_{2} | O_{1}\right]$, the conditional probability that the second card is even given that the first card is odd? (d) What is the conditional probability that the second card is odd given that the first card is odd?

Solution

VerifiedSince we have 3 cards, we have 6 possible combinations if the deck is dealt. Hence, Sample Space, S = {234, 243, 324, 342, 423,432}

Since each of the combination is equally likely, P(234) = P(243) = P(324) = P(342) = P(423) = P(432) = $\dfrac{1}{6}$

Now, We have $E_{1}$ = {234, 243, 423, 432} where first card is even numbered.

so, P($E_{1}$) = P(234) + P(243) + P(423) + P(432)

so, P($E_{1}$) = $\dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{3}$

And, $E_{2}$ = {243, 324, 342, 423} where second card dealt is even

so, P($E_{2}$) = P(243) + P(324) + P(342) + P(423)

so, P($E_{2}$) = $\dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{3}$

And, $E_{3}$ = {234, 324, 342, 432} where third card dealt is even

so, P($E_{3}$) = P(234) + P(324) + P(342) + P(432)

so, P($E_{3}$) = $\dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{3}$

Similarly, we have $O_{1}$ = {324, 342} where first card dealt is odd

so, P($O_{1}$) = P(324) + P(342) = $\dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{3}$

And, $O_{2}$ = {234, 432} where second card dealt is odd

P($O_{2}$) = P(234) + P(432) = $\dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{3}$

And, $O_{3}$ = {423, 243} where third card dealt is odd

P($O_{3}$) = P(423) + P(243) = $\dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{3}$

$\textbf{(a)}$

If we see $E_{1}$ and $E_{2}$, we find that $E_{1} \cap E_{2}$ = {243, 432}

Hence, P($E_{1} \cap E_{2}$) = P(243) + P(432) = $\dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{3}$

So, P($E_{2} | E_{1}$) = $\dfrac{P(E_{1} \cap E_{2})}{P(E_{1})}$ =$\dfrac{\frac{1}{3}}{\frac{2}{3}}$ = $\boxed{0.5}$