Question

Suppose we have 3 cards identical in form except that both sides of the first card are colored red, both sides of the second card are colored black, and one side of the third card is colored red and the other side is colored black. The 3 cards are mixed up in a hat, and 1 card is randomly selected and put down on the ground. If the upper side of the chosen card is colored red, what is the probability that the other side is colored black

Answer 1

Answer:

probability that the other side is colored black if the upper side of the chosen card is colored red = 1/3

Step-by-step explanation:

First of all;

Let B1 be the event that the card with two red sides is selected

Let B2 be the event that the

card with two black sides is selected

Let B3 be the event that the card with one red side and one black side is

selected

Let A be the event that the upper side of the selected card (when put down on the ground)

is red.

Now, from the question;

P(B3) = ⅓

P(A|B3) = ½

P(B1) = ⅓

P(A|B1) = 1

P(B2) = ⅓

P(A|B2)) = 0

(P(B3) = ⅓

P(A|B3) = ½

Now, we want to find the probability that the other side is colored black if the upper side of the chosen card is colored red. This probability is; P(B3|A). Thus, from the Bayes’ formula, it follows that;

P(B3|A) = [P(B3)•P(A|B3)]/[(P(B1)•P(A|B1)) + (P(B2)•P(A|B2)) + (P(B3)•P(A|B3))]

Thus;

P(B3|A) = [⅓×½]/[(⅓×1) + (⅓•0) + (⅓×½)]

P(B3|A) = (1/6)/(⅓ + 0 + 1/6)

P(B3|A) = (1/6)/(1/2)

P(B3|A) = 1/3