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# Basic Terms of Probability Section 3.2. Definitions Experiment: A process by which an observation or outcome is obtained. Sample Space: The set S of all.

## Presentation on theme: "Basic Terms of Probability Section 3.2. Definitions Experiment: A process by which an observation or outcome is obtained. Sample Space: The set S of all."— Presentation transcript:

Basic Terms of Probability Section 3.2

Definitions Experiment: A process by which an observation or outcome is obtained. Sample Space: The set S of all possible outcomes of an experiment. Event: Any subset E of the sample space S.

Probability of an Event Probability of an event is a measure of the likelihood that the event will occur. Remember probability is a number not a set. Mathematically speaking the probability of an event E denoted by P(E) is: P(E) = n(E)/n(S). Recall that that n(E) is the cardinal number of set E and n(S) is the cardinal number of set S.

Odds of an event Don’t confuse probability with odds. Every state lottery supposedly tells you the odds of winning. It turns out they don’t. They tell you the probability of winning. By definition, the odds of event E happening are denoted by o(E), which is given by: o(E) = n(E):n(E’). In words this says, that the odds of event E happening are number of times E happens divided by the number of times E does not happen. Think success compared with failure. Note odds use :, think of this as a fraction symbol or division sign.

NJ state Lottery According to the New Jersey state lottery, the odds of winning the Pick – 6 Jackpot is 1:13,983,816. However this is not correct. What they tell you is the probability. Remember the number of Pick – 6 numbers is 49 C 6. This number equals 13,983,816. The sample space contains 13,983,816 numbers and the winning number is just one of these. Hence a probability. The correct way to state the odds would be 1:13,983,815. There is 1 successful number and 13,983,815 unsuccessful number.

Relative frequency Tossing a single coin has a sample space of Heads and Tails. That is S={H,T}. Theoretically speaking the probability of tossing a head is ½. Let say you flip a coin 10 times and record the result of each toss. According to my results I recorded seven trials with heads. This would yield a relative frequency 7/10 or 0.7. We call this type of computation an empirical probability. When I toss the coin 100 times, my relative frequency was 52/100 or 0.52. This number is closer to the theoretical probability, but it is not exact. The relationship between relative frequencies and theoretical probabilities is called the Law of Large Numbers. This law states that if you repeat the experiment a large number of times, then the relative frequency of the outcome will tend to be close to the theoretical probability of that outcome.

Example 1 Find the probability and the odds of drawing the given cards 1. A black card 2. A heart 3. A two of clubs 4. A queen 5. A face card 6. A card above 9 (count ace as high)

Solution to example 1 Remember there are 52 cards in the deck. 26 are red, 26 are black. There are 4 suits diamonds (red), hearts (red), clubs(black), spades(black). There are 13 ranks in each suit 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A. The face cards are the J, Q, K. 1.The probability of drawing a black card is 26/52=1/2. The odds of drawing a black card is 26:26 or 1:1. 2.The probability of drawing a heart is 13/52=1/13. The odds of drawing a heart is 13:39=1:3. 3.The probability of drawing the two of clubs is 1/52. The odds of drawing the two of clubs is 1:51. 4.The probability of drawing a queen is 4/52=1/13. The odds of drawing a queen is 4:48=1:12. 5.The probability of drawing a face card is 12/52=3/13. The odds of drawing a face card is 12:40=3:10. 6.The odds of drawing a card above 9 is 20/52=5/13. The odds of drawing a face card is 20:32=5:8.

Example 2 Three coins are tossed answer the following: 1.What is the sample space? ANSWER S={HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}. Define the following events E is the event of tossing exactly two heads. F is the event that there are at least 2 are heads. G is the event that all three are heads. 2. What is P(E)? ANSWER n(E)/n(S)=3/8 3. What is O(E)? ANSWER n(E):n(E’)=3:5 4. What is P(F)? ANSWER n(F)/n(S)=4/8=1/2 5. What is O(F)? ANSWER n(F):n(F’)=4:4=1:1 6. What is P(G)? ANSWER n(G)/n(S)=1/8 7. What is O(G)? ANSWER n(G):n(G’)=1:7

Tossing 2 dice Consider tossing 2 six-sided dice. What is the sample space? We can enumerate the sample space by using ordered pairs like (x, y). X is the result of die 1 and y is the result of die 2. Below is the sample space (1,1), (1,2), (1,3), (1,4), (1,5), (1,6) (2,1), (2,2), (2,3), (2,4), (2,5), (2,6) (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) We see that there are 36 outcomes in this sample space. Suppose we add the faces together (which is what you do in most games that involve dice). What is the probability of tossing a 7? ANSWER: The outcomes that correspond to this event are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). There are 6 outcomes so the probability is 6/36.

More with two dice. Define E as the event of tossing 2 dice such that the sum of the faces is even. What is P(E)? ANSWER n(E)/n(S)=18/36=1/2. What is O(E)? ANSWER n(E):n(E’)=18:18=1:1. Define F as the event of tossing 2 dice such that the sum of the faces is greater than 9. What is P(F)? ANSWER n(F)/n(S)=6/36=1/6. What is O(F)? ANSWER n(F):n(F’)=6:30=1:5.

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