PSY 555 Homework 8

Answers

5.2(a). p(that you will win)=1/1000=.001

5.2(b). p(that your brother will win)=2/1000=.002

5.2(c). p(that one or the other of you will win)=.001+.002=.003

5.3(a). p(that you will win 2^nd prize given that you don’t win 1^st)=1/9=.111.

5.3(b). p(that he will win 1^st and you 2^nd)=(2/10)(1/9)

=(.20)(.111)=.022.

5.3(c). p(that you will win 1^st and he 2^nd)=(1/10)(2/9)

=(.10)(.22)=.022

5.3(d). p(that you are 1^st and he 2^nd [=.022])+p(that he is 1^st and you 2^nd [=.022])=p(that you and he will win 1^st and 2^nd)=.044

5.5. Conditional probabilities were involved in Exercise 5.3(a), as indicated by the ‘given that’ statement.

5.9. p(that the mother and child will look at each other at the same time during waking hours)=p(that mother looks at baby during waking hours)+p(that baby looks at mother during waking hours)=(2/13)(3/13)=(.154)(.231)=.036.

1. The additive law states that for mutually exclusive events, the probability of one or another event is the sum of the two events probabilities, p[A+B]=p(A)+p(B)

An example of the additive law would be the probability of getting a head or tail in a coin toss, p[head or tail]=p(head)+p(tail)=.5+.5=1.0.

The multiplicative law states that the probability of two or more independent events co-occuring is the product of all the events’ probabilities, p[A+B]=p(A)x p(B).

An example of the multiplicative law would be the probability of rolling two die and getting a 1 first, followed by a 3.

p[1 and 3]=p(1)x p(3)=(1/6)(1/6)=.028.

2. Joint probability is the likelihood of two events occurring at the same time. Conditional probability is the likelihood that an event will occur if another event has already occurred.

Joint probability example: the probability of a person being both male and blonde.

Conditional probability example: the probability that a person will call their insurance company given that they have been in a car accident.

3. For discrete variables, we can calculate the exact probability of a certain event. However, continuous variables require that we find the probability of an event within a specified interval.

4(a). There is a .2 probability that the person selected has O+ blood, p=(1/5)=.2.

4(b). There is a .248 probability that the person selected has Type O blood, p=(1/5)+(1/21)=.248.

4(c). There is .410 probability that the person selected has Type A blood, p=(1/3)+(1/13)=.410.

4(d). There is a .342 probability that the person selected has neither Type A nor Type O blood:

p=1-[(1/3)+(1/13)+(1/5)+(1/21)]=.342.

4(e). This is an impossible outcome, so the probability of this event is 0.

4(f). Blood types are mutually exclusive; you cannot have two types of blood.

5(a) There is a .27 probability that the person chosen will have none of the traits:

p=1-[.4+.25+.08]=1-.73=.27.

5(b). There is a .73 probability that the person chosen will have at least one trait, p=.4+.25+.08=.73.

5(c). There is a .33 probability that the person chosen will have

trait B, p=.25+.08=.33.

5(d). There is a .20 probability that the person chosen has both traits, given that the person has trait B: