An Introduction to…

chance likelihood 1

equally likely unlikely impossible very unlikely likely certain very likely

Probability of Events P(face card) ? 𝑃 𝐹𝐶 = 𝑛(𝐹𝐶) 𝑛(𝑆) 𝑃 𝐹𝐶 = 12 52 = 3 13

Rules of Probability Complement of an Event The complement of an event A is denoted by A′ and consists of everything in the sample space S except event A. 7

1 1 – P(E)

Rules of Probability Union of Two Events (Figure 5.5) The union of two events consists of all outcomes in the sample space S that are contained either in event A or in event B or both (denoted A  B or “A or B”).  may be read as “or” since one or the other or both events may occur. 10

Rules of Probability Intersection of Two Events The intersection of two events A and B (denoted A  B or “A and B”) is the event consisting of all outcomes in the sample space S that are contained in both event A and event B.  may be read as “and” since both events occur. This is a joint probability. 11

Example * First, draw a Venn diagram with Set A being even numbers, and Set B being prime numbers.

𝑃 𝐴 = 2 6 𝑃 𝐴′ = 4 6 𝑃 𝐴 +𝑃 𝐴′ = 6 6 𝑃 𝐴 +𝑃 𝐴′ =1 𝑃 𝐴′ =1−𝑃(𝐴) see next slide…

𝑃 𝐴 =0.7 𝑃 𝐵 =0.5 𝑃 𝐴∪𝐵 =0.9 𝑃 𝐴∩𝐵 =0.3 𝑃 𝐴 +𝑃 𝐵 =1.2 ? 𝑃 𝐴 +𝑃 𝐵 =1.2 ? * Impossible:  this is greater than 𝐴∪𝐵  total probability cannot exceed 1 𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃 𝐵 −𝑃 𝐴∩𝐵 CHECK: 0.9 =0.7+0.5−0.3

Rules of Probability General Law of Addition The general law of addition states that the probability of the union of two events A and B is: P(A  B) = P(A) + P(B) – P(A  B) So, you have to subtract P(A  B) to avoid over-stating the probability. When you add the P(A) and P(B) together, you count the P(A and B) twice. A and B A B

Events that are NOT Mutually Exclusive These events are not mutually exclusive because two of the 4 aces are also red cards so they would be counted TWICE if you added all the red cards + all the aces… Consider the problem of taking a card at random from a pack and you are wondering what is the chance you will get a red card or an ace?

Events that are NOT Mutually Exclusive P(red OR ace) = P(red) + P(ace) = 26/52 + 4/52 = 30/52 * This is INCORRECT ! What is the true probability of getting a red card OR an ace? P(red or ace) = 28/52

Practice ! You are now ready for some practice! Try page 354, Exercise 8J