Each of the possible five outcomes of a random experiment is equally likely. The sample space is {a, b, c, d, e}. All outcomes are equally likely. Let A denote the event {a, b}, and let B denote the event {c, d, e}. Determine the following probabilities:
a) P(A)
b) P(B)
c) P(A')
d) P(A∪B)
e) P(A∩B)

Respuesta :

Answer:

a) [tex] p(A) = \frac{2}{5}[/tex]

b) [tex] p(B) =\frac{3}{5}[/tex]

c) [tex] p(A') = 1-p(A) = 1-\frac{2}{5} = \frac{3}{5}[/tex]

d) The probability for intersection on this case is 0 because the sets A and B not have any element in common, so then we have this

[tex] P(AUB) = P(A) +P(B) -0 = \frac{2}{5} +\frac{3}{5} =1[/tex]

e) The intersection for this case is the empty set between the sets A and B so for this reason the probability is 0

P(A∩B)=0

Step-by-step explanation:

For this case we have the following sample space:

[tex] S= [a,b,c,d,e][/tex]

And we have defined the following events:

[tex] A= [a,b][/tex]

[tex] B= [c,d,e][/tex]

For this case we can find the probabilities for each event using the following definition of probability:

[tex] p =\frac{Possible cases}{total cases}[/tex]

The total cases for this case are 5 , the possible cass for A are and for B are 3.

Usign this we have this:

[tex] p(A) = \frac{2}{5}, p(B) = \frac{3}{5}[/tex]

Then we can find the following probabilites:

a) P(A)

[tex] p(A) = \frac{2}{5}[/tex]

b) P(B)

[tex] p(B) =\frac{3}{5}[/tex]

c) P(A')

Using the complement rule we have this:

[tex] p(A') = 1-p(A) = 1-\frac{2}{5} = \frac{3}{5}[/tex]

d) P(A∪B)

For this case we can use the total probability rule and we got:

[tex] P(AUB) = P(A) +P(B) -P(A and B)[/tex]

The probability for intersection on this case is 0 because the sets A and B not have any element in common, so then we have this

[tex] P(AUB) = P(A) +P(B) -0 = \frac{2}{5} +\frac{3}{5} =1[/tex]

e) P(A∩B)

The intersection for this case is the empty set between the sets A and B so for this reason the probability is 0

P(A∩B)=0