A fair die is tossed once. Let the event {A} is getting an even number, the event {B} - number 2 #>=# 3. How do you find P(A U B), P(A/B)? Are the events A, B dependent or not?

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Answer 1 · 2015-11-09T17:28:27

#P(AuuB)=5/6# #P(A//B)=1/2#
The events #A# and #B# are independent

To calculate probability in such tasks you have to write doown all elementary events first:

#Omega={1,2,3,4,5,6}#

#bar(bar(Omega))=6#

Event #A# is: "Getting an even number", so:

#A={2,4,6}#

#bar(bar(A))=3#

#P(A)=(bar(bar(A)))/(bar(bar(Omega)))=1/2#

Event #B# is: "Getting a number greater than or equal to 3", so

#B={3,4,5,6}#

#bar(bar(B))=4#

#P(B)=4/6=2/3#

Now we have to write what is #AuuB#

The operator #uu# for events means "or" so the event #AuuB# means "getting an even number or number greater than or equal to #3#"

So #AuuB={2,4,6,3,5}#, #bar(bar(AuuB))=5#

To calculate conditional probability #P(A//B)# we need to find probrbility #P(AnnB)#

#AnnB# means "Tossing an even number and number greater than or equal to 3",
so #AnnB={4,6}#, #bar(bar(AnnB))=2#

Now we can calculate required probabilities:

#P(AuuB)=bar(bar(AuuB))/bar(bar(Omega))=5/6#

#P(AnnB)=bar(bar(AnnB))/bar(bar(Omega))=2/6=1/3#

To calculate #P(A//B)# we use the formula:

#P(A//B)=(P(AnnB))/(P(B))=1/3:2/3=1/3*3/2=1/2#

The events #A# and #B# are called independent if and only if
#P(AnnB)=P(A)*P(B)#

For these events we have:

#P(AnnB)=1/3#

#P(A)*P(B)=1/2*2/3=1/3#

Since #P(A)*P(B)# and #P(AnnB)# are equal we can write, that #A# and #B# are independent events.