5.2 STATISTICS PROBABILITIES MARCH 30
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
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. )
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.