Question

You have a shuffled deck of three cards: 2, 3, and 4, and you deal out the three cards. Let EiE_{i} denote the event that ith card dealt is even numbered. (a) What is P[E2E1]\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 OiO_{i} represent the event that the ith card dealt is odd numbered. What is P[E2O1]\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

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Since 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) = 16\dfrac{1}{6}

Now, We have E1E_{1} = {234, 243, 423, 432} where first card is even numbered.

so, P(E1E_{1}) = P(234) + P(243) + P(423) + P(432)

so, P(E1E_{1}) = 16+16+16+16=23\dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{3}

And, E2E_{2} = {243, 324, 342, 423} where second card dealt is even

so, P(E2E_{2}) = P(243) + P(324) + P(342) + P(423)

so, P(E2E_{2}) = 16+16+16+16=23\dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{3}

And, E3E_{3} = {234, 324, 342, 432} where third card dealt is even

so, P(E3E_{3}) = P(234) + P(324) + P(342) + P(432)

so, P(E3E_{3}) = 16+16+16+16=23\dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{3}

Similarly, we have O1O_{1} = {324, 342} where first card dealt is odd

so, P(O1O_{1}) = P(324) + P(342) = 16+16=13\dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{3}

And, O2O_{2} = {234, 432} where second card dealt is odd

P(O2O_{2}) = P(234) + P(432) = 16+16=13\dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{3}

And, O3O_{3} = {423, 243} where third card dealt is odd

P(O3O_{3}) = P(423) + P(243) = 16+16=13\dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{3}

(a)\textbf{(a)}

If we see E1E_{1} and E2E_{2}, we find that E1E2E_{1} \cap E_{2} = {243, 432}

Hence, P(E1E2E_{1} \cap E_{2}) = P(243) + P(432) = 16+16=13\dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{3}

So, P(E2E1E_{2} | E_{1}) = P(E1E2)P(E1)\dfrac{P(E_{1} \cap E_{2})}{P(E_{1})} =1323\dfrac{\frac{1}{3}}{\frac{2}{3}} = 0.5\boxed{0.5}

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