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P_joint (x_i, y_k) = P_coin (x_i) times P_+die(y_k) Equation 3.15 is correct eve

ID: 3200490 • Letter: P

Question

P_joint (x_i, y_k) = P_coin (x_i) times P_+die(y_k) Equation 3.15 is correct even for loaded dice (the P_die(y_k) aren't all equal to 1/6) or a two-headed coin (P_coin(x_1) = 1, P_coin(x_2) = 0). On the other hand, for two connected events (for example, the chance of rain versus the chance of hail), we don t get such a simple relation. Show that if P_coin and P_die are correctly normalized, then so will be P_joint- Suppose we roll two dice. What's the probability that the numbers on the dice add up to 2? To 6? To 12? Think about how you used both the addition and the multiplication rule for this.

Explanation / Answer

total number of possibilities = 36

sample space such that add ups to 2

= { (1,1),(1,1)}

probability = 2/36 = 1/18

such adds up to 6

sample space = {(1,5),(2,4),(3,3),(4,2),(5,1)}

probability = 5/36

add ups to 12

sample space = {(6,6),(6,6)}

probability = 2/36 = 1/18