Need your help in probability problems? Thanks :)

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2 Answers
Apr 29, 2018

See below

Explanation:

P(x)=#8/12=2/3#

P(Y)=#5/12#

P(X nn Y)#=2/12=1/6# the numbers 1 and 3 are in both events.

P(X uu Y)#=11/12# the number 12 is not in either event

The sample space would be the numbers 1 to 12

The events are dependent as 1 and 3 are in both events

Given event A has occurred P(even)=#2/8=1/4#
Sorry don't know what P(X/Y) means

Apr 29, 2018

What is #"P"(X)#?

The probability of event #X# occurring is

#"P"(X)="the number of elements in X"/"the number of elements in S"#

Since there are 12 possible rolls, #S# has 12 elements. The event #X# contains 8 of those 12. So, #"P"(X)# is

#"P"(X)=8/12#

#color(white)("P"(X)) = 2/3#

What is #"P"(Y)#?

Similarly, #Y# contains 5 of the 12 possible rolls, so

#"P"(Y)=5/12#

What is #"P"(X nn Y)#?

The event #XnnY# is the set of elements that are in both #X# and #Y#. By observation, we see that #XnnY={1,3}.# Thus, the size of #XnnY# is 2, and the probability of #XnnY# is

#"P"(XnnY) = 2/12#

#color(white)("P"(XnnY)) = 1/6#

What is #"P"(XuuY)#?

The event #XuuY# is the set of elements that are in #X# or #Y#. If an element appears in at least one of #X# or #Y#, then it is in #XuuY#.

By observation, we see #XuuY={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}#. (The only element that does not appear in either #X# or #Y# is #{12}#.) Thus the probability of #XuuY# is

#"P"(XuuY) = 11/12#

Write down the sample space #S# of possible outcomes.

#S# is the event that anything happens. It contains all the elements that we could possibly observe, which makes #"P"(S)=100%.# For a standard 12-sided die, the sample space of a single roll is

#S={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}#

Are #X# and #Y# independent? Explain.

Two events #X# and #Y# are independent if

#"P"(XnnY) = "P"(X) xx "P"(Y)#

From above, we know #"P"(XnnY)=1/6#. We also know #"P"(X)=2/3# and #"P"(Y)=5/12#. We plug these values into the equation to check:

#1/6 stackrel"? "= 2/3 xx 5/12=(1xx5)/(3xx6)=5/18 != 1/6#

Since #"P"(XnnY) != "P"(X) xx "P"(Y)#, we know #X# and #Y# are not independent.

What is the probability that you rolled an even number, given that event #A# has occurred?

There is no event #A# defined in the question. Assuming they mean event #X#, we seek the ratio

#"P"("even"|X)= "the number of even numbers in X"/"the size of X"#

Event #X# has 2 even numbers in it: #{4, 8}#. So the probability of rolling an even number, given that #X# has occurred, is:

#"P"("even"|X)= 2/8#

#color(white)("P"("even"|X))= 1/4#

What is #"P"(X|Y)#?

Similarly, #"P"(X|Y)# is the probability of rolling a number that's also in #X#, given that it's known to be in #Y#.

#"P"(X|Y) =("the size of X "nn" Y")/"the size of Y"= ("P"(X nn Y))/("P"(Y))#

#color(white)("P"(X|Y)) = 2/5#