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:

_{}