Probability of event A or B occuring given 3 events A, B and C

Question

I am struggling a little bit with Probability.

I am given 3 events after tossing 2 triangular die. ( 4 sides, numbered 1 to 4))

Event A: the bottom faces sum up to 6 (3/16)

Event B: each bottom face shows an even number (4/16)

Event C: each bottom face shows the same number(doubles) (4/16)

I am asked to find P(A U B)

So the formula I was given for Union was P(A U B) = P(A) + P(B) - P(B ^ A)..

So I did P(A U B) = (3/16) + (4/16) - (3/16 * 4/16)

BUT this is not the right answer. Do I also have to account for C? I.e

P(A U B) = P(A) + P(B) - P(C) -P(A ^ B) - P(A ^ C) - P(B ^ C)

Is this correct? ( I have a limited number of tries on the problem)

Answer 1

The outcome space consists of $16$ atoms occuring without bias; the ordered pairs of results for the two, fair tetrahedral-die.

Event A: the bottom faces sum up to 6 (3/16)

$A=\{(2,4),(3,3),(4,2)\}$ so yes.

Event B: each bottom face shows an even number (4/16)

$B=\{(2,2),(2,4),(4,2),(4,4)\}$ so yes.

Event C: each bottom face shows the same number(doubles) (4/16)

$C=\{(1,1),(2,2),(3,3),(4,4)\}$ so yes.

I am asked to find $P(A \cup B)$

What is the set $A\cup B$? How many atoms does it hold? That will immediately tell you the answer.

So the formula I was given for Union was $P(A \cup B) = P(A) + P(B) - P(B \cap A)$..

Well, sure, that is so.   However, what is $A\cap B$?   How many atoms are in this intersection?   That will lead you to the answer.

So I did $P(A \cup B) = (3/16) + (4/16) - (3/16 * 4/16)$

BUT this is not the right answer.

Indeed it is not.   You are using the product rule for the probability for an intersection of independent events.   They are not independent.