Divine

Divina Sison 11-12 am/MWF Prof. Tabuloc

Conditional Probability 1. An Experiment consists of tossing a coin and a die. If E1 is the event that “head” comes up tossing the coin and E2 is the event that ‘3 or 6’ cones up in tossing the die, state in words the meaning of each of the ff: a. b. c. d.

E1E2 Pr{E1|E2} E1 E2

Answer: a. Heads on the coin and 3 or 6 on the die. b. Probability of heads on the coin, given that a 3 or 6 has come up on the die. c. Tails on the coin and anything on the die. d. 1, 2, 4, or 5 on the die and anything on the coin.

Dependent events Suppose that a box contains 3 white balls and 2 black balls. Let E1 be the event “first ball drawn is black” and E2 the event “second ball drawn is black”, where the balls are not replaced after being drawn. Here E1 and E2 are dependent events. The probability that the first ball drawn is black is Pr {E1} = 2/ (3+2) =2/5. The probability that the second ball drawn is black, given that the first ball drawn was black ,is Pr{E2| E1}=1/(3+1)=1/4.Thus the probability that both balls drawn are black is Pr {E1E2} =Pr {E1} Pr {E2|E1} =2/5 .1/4=1/10

Independent events Let E1 and E2 be the events “heads on fifth toss” and “heads on sixth toss” of a coin, respectively. Then E1 and E2 are independent events and thus the probability of heads on both the fifth and sixth tosses (assuming the coin to be fair) Pr {E1E2} =Pr {E1} Pr {E2} = (½)(½)=(¼)

Mutually Exclusive events If E1 Is the event “drawing an ace from a deck of cards” and E2 is the event “drawing a king”, then Pr {E1}=4/52=1/3 and Pr{E2}=4/52=1/3. The probability of drawing either an ace or a king in a single draw is Pr {E1 + E2} =Pr {E1} + Pr {E2} =1/3 + 1/13=2/13

Non-Mutually events If E1 is the event “drawing an ace” from a deck of cards and E2 is the event “drawing a spade”, then E1 and E2 are not mutually exclusive since the ace of spades can be drawn. Thus the probability of drawing either an ace or a spade or both is Pr {E1 + E2} =Pr {E1} + Pr {E2}-Pr {E1E2} =4/52+13/52-1/52=16/52=4/13

PRODUCT SETS Prove: A X (B C) = (AXB) (A X C) A X (B C) = {(x, y): X A, y B C} = {(x, y): x A, y B, y C} = {(X, y): (x, y) A x B, (x, y) A x C} = (A x B) (A x C)