Probability

1. Probability Presented by Tutorial Services The Math Center

2. Introduction to Probability • Probability of an event.Given an event E, the probability of the event, Pr(E), can be found by: n(E) • n(E) is the number of occurrences of the event. • n(S) is the number of elements in the sample space. Pr(E) = n(S)

3. Empirical probability.Pr(event) = Number of occurrences of event Number of total trials or observations • Example: A coin was flipped four times. It showed heads on the first flip, tails on the second, heads on the third, and heads on the fourth. Find the empirical probability of getting tails. Solution: Number of occurrences of tails 1 Pr(Tails) = = Total number of trials 4

4. S E’ E Unions and Intersections • Pr (E and F both occur) = Pr (E ∩ F) • Pr (E or F both occur) = Pr (E U F) • Pr (E does not occur) = Pr (E’) F E

5. If E and F are two inclusive events, the probability that one event or the other will occur is given by Pr (E or F) = Pr (E U F) = Pr (E) + Pr (F) – Pr (E ∩ F) 3 1 4 9 9 9 Inclusive Events E ∩ F Example: A card is drawn from a deck of nine cards numbered 1 through 9. What is the probability that it is even or divisible by 3? Solution: Pr (even)= Pr (divisible by 3) = Pr (even & divisible by 3)= . Pr ( even or divisible by 3) = _4_ + _3_ - _1_ = _6_ = _2_ 9 9 9 9 3

6. Pr (E ∩ F) = 0, and Pr (E or F) = Pr (EUF) = Pr (E) + Pr (F) For mutually exclusive events E1, E2,…, En Pr (E1U E2U …U En) = Pr(E1) + Pr(E2) + … Pr (En) . S E F Mutually Exclusive Events

7. 98 500 72 500 • Example: 500 students apply for a scholarship. 98 were students who were first in their class and 72 were second in their class. What is the probability that a student chosen at random is first in their class or second in their class? • Solution: Let E be the event “choosing first in class ” and F be the event “choosing second in class”. Pr (E) = Pr (F) = Pr ( E U F) = Pr (E) + Pr (F) = _98_ + _72_ = _170_ = _17_ 500 500 500 50

8. Conditional Probability • The Probability that A occurs, given that B occurs is given by, Pr (A | B) = n (A ∩ B)n (B) • Let A and B be events in the sample space S with Pr (B) > 0. The conditional probability that event A occurs, given that event B occurs is given by, Pr (A | B) = Pr (A ∩ B) Pr (B)

9. 4 36 4 36 3 36 2 36 1 36 10 36 • ExampleA pair of 6-sided die is rolled. What is the probability that the sum rolled on the die is 9 given that the sum is greater than 8? • Solution: Pr(sum is 9 ∩ sum >8) =Pr(sum is 9) = Pr(Sum > 8) = + + + = Pr(sum is 9 |sum >8) = 4/ 36 = 4 36 Pr (A | B) = Pr (A ∩ B) Pr (B) . 36 10 10/36 4 2 = = 10 5

10. Product Rule for Events • If A and B are probability events, then the probability of the event “A and B” can be found by using: i. Pr (A and B) = Pr (A ∩ B) = Pr (A) · Pr (B|A) ii. Pr (A and B) = Pr (A ∩ B) = Pr (B) · Pr (A|B) • If A and B are independent events then: Pr(A and B) = Pr(A ∩ B) = Pr(A) · Pr(B)

11. 12 52 1 2 Example: Independent Events A card is pulled from a deck of 52 cards and a coin is tossed. Find the probability of pulling a face card from the deck (Jack, Queen, and King) and getting a tails on the coin. Solution: Pr(A) = Pr(pulling a face card) = Pr(B) = Pr(tails on coin) = Pr(A ∩ B) = Pr(A) · Pr(B) = 12· 1 = 6 = 3 52 2 52 26

12. Probability Tree • Draw a probability tree when you have sequential occurrences. • The probability of the event occurring is the product of the probabilities on the one particular branch. • If an event can be described by two or more paths through a probability tree, its probability is found by adding the probabilities from the paths. Product of branch probabilities on path leading to F1through E1 Pr(E1| F1) = Sum of all branch products on paths leading to F1

13. Example: The Queen’s jester had made a joke at the expense of the Queen; usually punished by death, but the Queen liked this particular jester. She was going to give him a chance to live. She gave the jester 5 black balls, 5 white balls, and two urns and told him, “Put the balls in either urn as you feel. I will flip a coin and pick an urn to pull a ball from. If the ball I pull is black, then you will die. If it’s white, you will live.” If the Jester puts one white ball in the first urn and the rest in the second urn, what is the probability that she chooses the first Urn given that the ball he gets is white ? 1 White Ball Urn 1 1/2 White Ball 4 9 5 9 Urn 2 1/2 9 13 Black Ball Solution: (1/2)· (1) Pr(U1|W) = (1/2) · (1) + (1/2) · (4/9) =

14. Bayes’ Formula • If a probability experiment has n possible outcomes in its first stage given by E1, E2,…, En, and if F1is an event in the second stage, then the probability that event E1 occurs in the first stage, given that F1has occurred in the second, is Pr(E1| F1) = Pr(E1) · Pr(F1| E1) Pr(E1) · Pr(F1| E1) + Pr(E2) · Pr(F1| E2) …+ Pr(En) · Pr(F1 | En)

15. The Queen’s jester had made a joke at the expense of the Queen; usually punished by death, but the Queen liked this particular jester. She was going to give him a chance to live. She gave the jester 5 black balls, 5 white balls, and two urns and told him, “Put the balls in either urn as you feel. I will flip a coin and pick an urn to pull a ball from. If the ball I pull is black, then you will die. If it’s white, you will live.” If the Jester puts one white ball in the first urn and the rest in the second urn, what is the probability that she chooses the first Urn given that the ball he gets is white ? Solution: Pr(U1) = 1/2 Pr(U2) = 1/2 Pr(W | U1) = 1 Pr(W | U2) = 4/9 Pr (U1) · Pr(W |U1)_____ _ Pr (U1) · Pr(W |U1) +Pr (U2) · Pr(W |U2) Bayes’ Formula (1/2)· (1) = (1/2)· (1) + (1/2)· (4/9) = 1/2 = 9 13/18 13

16. Links and Handouts • Probability Handout • Probability Quiz