Questions

2-1. Suppose you have a deck with only three cards. Each card has two sides, and each side is either black or white. One card has two black sides. The second card has one black and one white side. The third card has two white sides. Now suppose all three cards are placed in a bag and shuffled. Someone reaches into the bag and pulls out a card and places it flat on a table. A black side is shown facing up, but you don’t know the color of the side facing down. Show that the probability that the other side is also black is 2/3. Use the counting method (Section 2 of the chapter) to approach this problem. This means counting up the ways that each card could produce the observed data (a black side facing up on the table).

Answer: Use \(X\), \(Y\) and \(Z\) to denote the event of drawing first, second, third card respectively.Use \(B\) and \(W\) to denote the event of drawing black or white facing up respectively. Since the card drawn has at least one side that’s black, we know the only chance the other side is black is the card is event \(X\), and it cannot be event \(Z\) since it has no black side. \(P(X|B) = \frac{P(X \cap B)}{P(B)} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{2}{3}\).

Counting method: situation could be \(XB1\), \(XB2\) or \(YB\), so \(\frac{2}{3}\).

p_x_and_b = 1/2
p_b = 3/4
p_x_given_b = p_x_and_b / p_b
p_x_given_b
## [1] 0.6666667

2-2. Now suppose there are four cards: B/B, B/W, W/W, and another B/B. Again suppose a card is drawn from the bag and a black side appears face up. Again calculate the probability that the other side is black.

Answer: Use \(W\) to denote the new event of drawing the fourth card. \(P(X|B) + P(W/B) = P(X|B) \cdot 2 = \frac{P(X \cap B) \cdot 2}{P(B)} = \frac{\frac{1}{3} \cdot 2}{\frac{5}{6}} = \frac{4}{5}\)

p_x_and_b = 1/3
p_b = 5/6
p_x_given_b = p_x_and_b / p_b
p_x_given_b
## [1] 0.4

2-3. Imagine that black ink is heavy, and so cards with black sides are heavier than cards with white sides. As a result, it’s less likely that a card with black sides is pulled from the bag. So again assume there are three cards: B/B, B/W, and W/W. After experimenting a number of times, you conclude that for every way to pull the B/B card from the bag, there are 2 ways to pull the B/W card and 3 ways to pull the W/W card. Again suppose that a card is pulled and a black side appears face up. Show that the probability the other side is black is now 0.5. Use the counting method, as before.

Answer: \(P(X|B) = \frac{P(X \cap B)}{P(B)} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}\).

Counting method: situation could be \(XB1\), \(XB2\) or \(YB\). But \(YB\) has double the probability, so \(2\) out of \(4\).

p_x_and_b = 1/3
p_b = 2/3
p_x_given_b = p_x_and_b / p_b
p_x_given_b
## [1] 0.5

2-4. Assume again the original card problem, with a single card showing a black side face up. Before looking at the other side, we draw another card from the bag and lay it face up on the table. The face that is shown on the new card is white. Show that the probability that the first card, the one showing a black side, has black on its other side is now 0.75. Use the counting method, if you can. Hint: Treat this like the sequence of globe tosses, counting all the ways to see each observation, for each possible first card.

Answer: Original problem could be \(XB1\), \(XB2\) or \(YB\). Now there is a second event, so possibilities are: \(XB1,YW\), \(XB1,ZW1\), \(XB1,ZW2\),\(XB2,YW\), \(XB2,ZW1\), \(XB2,ZW2\),\(YB,ZW1\), \(YB,ZW2\). So \(\frac{6}{8} = \frac{3}{4}\).

2-5. Suppose there are two species of panda bear. Both are equally common in the wild and live in the same places. They look exactly alike and eat the same food, and there is yet no genetic assay capable of telling them apart. They differ however in their family sizes. Species A gives birth to twins 10% of the time, otherwise birthing a single infant. Species B births twins 20% of the time, otherwise birthing singleton infants. Assume these numbers are known with certainty, from many years of field research. Now suppose you are managing a captive panda breeding program. You have a new female panda of unknown species, and she has just given birth to twins. What is the probability that her next birth will also be twins?

Answer: Use \(A\) and \(B\) to denote probability of it being species A and B respectively, use \(S\) and \(T\) to denote probability of giving birth to single infant vs. twins. We know \(P(A) = P(B) = 0.5\), \(P(T|A) = 0.1\), \(P(T|B) = 0.2\).

Then probability of next twin birth is \(P(A|T) \cdot P(T|A) + P(B|T) \cdot P(T|B)\). \(P(A|T) = \frac{P(A \cap T)}{P(T)} = \frac{0.5 \cdot 0.1}{0.15} = \frac{1}{3}\), \(P(B|T) = 1 - P(A|T) = \frac{2}{3}\). So the answer is \(\frac{1}{3} \cdot 0.1 + \frac{2}{3} \cdot 0.2 = \frac{1}{6}\)