MHB Probability Question with Mutually Exclusive and Indpendent Events

[cmkluza]

Hello, I'm sorry for the title of this thread, but I don't actually even know what to call this. I'm very bad at probability and statistics in general, but I have the following problem:

Events A and B are such that P(A) = 0.3 and P(B) = 0.4.
(a) Find the value of P(A ∪ B) when
(i) A and B are mutually exclusive;
(ii) A and B are independent.

(b) Given that P(A ∪ B) = 0.6, find P(A | B).

I would like to start trying this, but I don't really know where to start. I can hardly understand the wording used in these kinds of problems, much less make sense of the math. Can anyone make any suggestions to help me out?

Any help is greatly appreciated!

 

[SuperSonic4]

Hello, I'm sorry for the title of this thread, but I don't actually even know what to call this. I'm very bad at probability and statistics in general, but I have the following problem:

Events A and B are such that P(A) = 0.3 and P(B) = 0.4.
(a) Find the value of P(A ∪ B) when
(i) A and B are mutually exclusive;
(ii) A and B are independent.

(b) Given that P(A ∪ B) = 0.6, find P(A | B).

I would like to start trying this, but I don't really know where to start. I can hardly understand the wording used in these kinds of problems, much less make sense of the math. Can anyone make any suggestions to help me out?

Any help is greatly appreciated!

Your question is fine :).
For a conceptual idea you can think of tossing a fair coin: A is heads and B is tails.

Are you aware of any formulas for probabilities? In particular $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Notation:

• [math]P(A \cap B)[/math] is the intersection but I think of it as and (as in getting heads and tails)
  
• [math]P(A \cup B)[/math] is called the union but I find it easier to think of it as or (getting heads or tails)

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If events A and B are mutually exclusive it means that the chance of A and B happening is 0. In our coin example you cannot get heads and tails so they are mutually exclusive.

Using notation this is: $$P(A \cap B) = 0$$

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Independent events are ones which do not change however many times you do them. Tossing a coin will always be 50/50 whether it's the first or the fiftieth toss. Contrast drawing cards from a deck (without replacement): if you draw the 3 of hearts you cannot draw it a second time.

In this case: [math]P(A \cap B) = P(A)P(B)[/math]

Can you apply this to your problem?