5.2 STATISTICS PROBABILITIES MARCH 30

2. 5.2.11 A probability experiment is conducted in which the sample space of the experiment is S={6,7,8,9,10,11,12,13,14,15,16,17} = (12#s) Let event E={6,7,8,9}. Assume each outcome is equally likely. List the outcomes in Ec. Find P (Ec) .

1. List the outcomes in Ec= # are in S but not in E ◙ Ec= {10,11,12,13,14,15,16,17} = 8 2. P (Ec) = ♦Ec or N(Ec).= non common #s = 8 ♦S or N(S) = all #{}= 12 P (Ec) = N(Ec)/N(S) P (Ec) = 8/12 = 0.667 .

4. 5.2.23 A golf ball is selected at random from a golf bag. If the golf bag contains 8 type A​ balls, 6 type B​ balls, and 4 type C​ balls, find the probability that the golf ball is not a type A ball.

A 1. Let S denote the sample space of a probability experiment and let E denote an event. 2. The complement of​ E, denoted Ec​, is all outcomes in the sample space S that are not outcomes in the event E. ◙ Note that the event that the golf ball is not a type A ball is the complement of the event that the golf ball is a type A ball. 3. Let A be the event that the golf ball is a type A ball. Then Ac is the event that the selected golf ball is not a type A ball. 4.If S is the sample space of this​ experiment, the probability of A is given by the​ formula, where​ N(A) is the number of outcomes in​ A, and​ N(S) is the number of outcomes in the sample space. P(A)=N(A)/N(S) ◙ Find the number of outcomes in A. N(A) = 8 ◙ Find​ N(S), the number of outcomes in the sample space. N(S)= 8+6+4 = 18 P(A)=N(A)/N(S) P(A)=8/18 P(A) =0.444 B If E represents any event and Ec represents the Complement of​ E, then the probability P(Ec) is given by the formula below. P(EC)=1-P(E) Now find P(Ac)​, the probability that the selected golf ball is not a type A golf ball. P(Ac)​= 1-P(A)= 1-0.444= 0.556

1. 5.2.5 A probability experiment is conducted in which the sample space of the experiment is S= {9,10,11,12,13,14,15,16,17,18,19,20}. Let event E={10,11,12,13,14,15} and event F={14,15,16,17}. List the outcomes in E and F. Are E and F mutually​ exclusive?

HELP ME TO SOLVE THIS... The list of outcomes in E and F are the outcomes that event E and event F have in common. ►Notice that the outcomes in event E are ​{10,11,12,13,14,15​} and the outcomes in event F are {14,15,16,17}. Are there any outcomes in common in both E and​ F? YES ◙ There are outcomes in common in E and F.​ Thus, what is the correct list of outcomes in E and​ F? {14,15 } ◙Two events are disjoint if they have no Outcomes in common. Another name for disjoint events is mutually exclusive events. ​Thus, are E and F mutually​ exclusive? ◙ No. E and F have outcomes in common.

2. 5.2.11 A probability experiment is conducted in which the sample space of the experiment is S={4,5,6,7,8,9,10,11,12,13,14,15}.= (12 = N(S) = 12) Let event E={5,6,7,8,9,10,11,12}. Assume each outcome is equally likely. List the outcomes in Ec. Find P (Ec) ◙ List the outcomes in Ec.

The complement of​ E, denoted Ec​, is all outcomes in the sample space S that are not outcomes in the event E. ► Notice that the outcomes in the sample space S are given in the problem statement along with the outcomes in event E. Are there outcomes in the sample space S that are not in event​ E? YES ◙ What outcomes are listed in the sample space S that are not listed in event​ E? ◙ The outcomes are {4,13,14,15} Thus, the outcomes in Ec are {4,13,14,15}.= 4= N(Ec). To find P(Ec)​, divide the number of outcomes in Ec by the number of outcomes in S. ►First find N(Ec). N(Ec).= 4 ►Now find​ N(S). ​N(S) = 12 ► Now find P(Ec). P(Ec) = N(Ec)/ N(S). P(Ec) = 4/12 = 0.333

4. 5.2.23 A golf ball is selected at random from a golf bag. If the golf bag contains 6 type A​ balls, 9 type B​ balls, and 5 type C​ balls, find the probability that the golf ball is not a type A ball.

The probability that the golf ball is not a type A ball is N(A) = 6 N(S) = 6+9+5 N(S) = 20 P(A)=N(A)/N(S) P(A)= 6/20 P(A)=0.3 (event golf ball is a type A ball. ) P(Ac)​= 1-P(A)= = 1-0.3 = 0.7 (event golf ball is NOT a type A ball. )

3. 5.2.13 Find the probability of the indicated event if ​P(E)=0.30 and ​P(F)=0.50. Find​ P(E or​ F) if​ P(E and ​F)=0.20.

Use the general addition rule to find the probability. The rule states that for any two events E and​ F, 1.P(E or ​F)=​P(E)+​P(F)−​P(E and​ F). 2. Substitute the values from the problem statement into the general addition rule. ​P(E or​ F) = ​P(E)+​P(F)−​P(E and​ F) P(E or​ F) =0.30+0.50-0.20 P(E or​ F) = 0.6

1. 5.2.5 A probability experiment is conducted in which the sample space of the experiment is S= {9,10,11,12,13,14,15,16,17,18,19,20,21,22}. Let event E={12,13,14,15,16,17} and event F={16,17,18,19}. List the outcomes in E and F. Are E and F mutually​ exclusive? .

◙ List the outcomes in E and F {16,17} (#s in both are common) ◙Are E and F mutually​ exclusive? ◙No. E and F have outcomes in common.