A pair of fair dice is tossed.  Let A = {the sum is 7} and B = {3 appears on at least one die}Find each of the following.  Write your answer as a fraction.P(A) = P(B) = P(A∩B)=P(B | A) = P(A | B) = 

A pair of fair dice is tossed Let A the sum is 7 and B 3 appears on at least one dieFind each of the following Write your answer as a fractionPA PB PABPB A PA B class=

Respuesta :

To find the probability we need to count each one of the cases.

The event A is the sum of the dices are 7, since the dices are distinguishable between them we have that the event A is the set:

[tex]A=\lbrace(6,1),(5,2),(4,3),(3.4),(2.5),(1,6)\}[/tex]

Now the event B is the set:

[tex]B=\lbrace(3,1),(3.2),(3,3),(3,4),(3,5),(3,6),(1,3),(2.3),(4,3),(5,3),(6,3)\}[/tex]

Now that we have our sets A and B we can calculate the other ones:

[tex]A\cap B=\lbrace(4,3),(3,4)\}[/tex][tex]B|A=\lbrace(3,1),(3,2),(3,3),(3,5),(3,6),(1,3),(2,3),(5,3),(6,3)\}[/tex][tex]A|B=\lbrace(6,1),(5,2),(2,5),(1,6)\}[/tex]

Now that we have all our sets we have to count how many cases each of them have and divided them by the total number os possible outcomes (36).

Then:

[tex]P(A)=\frac{6}{36}=\frac{1}{6}[/tex][tex]P(B)=\frac{11}{36}[/tex][tex]P(A\cap B)=\frac{2}{36}=\frac{1}{18}[/tex][tex]P(B|A)=\frac{9}{36}=\frac{1}{4}[/tex][tex]P(A|B)=\frac{4}{36}=\frac{1}{9}[/tex]

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