Probability.

[winniethepooh]

A fair die is rolled once ana a fair coin is flipped once. What is the probability that either the die will land on 3 or that the coin will land on heads?

[Ashley@VeritasPrep]

Hey there,

The quickest way to do a problem like this is to add the separate probabilities (of a 3 and of heads) and subtract the overlap once, since we'll have counted that overlap twice in adding. The probability of getting a 3 is 1/6 and the probability of getting heads is 1/2, but that 1/6 includes both head-getting and tail-getting (and the 1/2 includes both 3-getting and non-3-getting). We will get heads 1/2 of the time that we get a 3, so in other words 1/2 * 1/6 of the time. There's your area of overlap.

So your equation becomes 1/6 + 1/2 - (1/6)(1/2). (Note that this method still counts that intersection in which we get a 3 AND heads, but it properly counts it just once instead of twice. If we had simply added 1/6 and 1/2, we would have been counting the intersection twice.)

Does that make sense? If not, I'll clarify with diagrams

[winniethepooh]

It does make sense, But I didn't get the perfect picture of the solution!

[Ian Stewart]

A fair die is rolled once ana a fair coin is flipped once. What is the probability that either the die will land on 3 or that the coin will land on heads?

The probability that we don't get a 3 on the die is 5/6, and the probability we don't get Heads is 1/2, so the probability neither event happens is (1/2)(5/6) = 5/12. Thus the probability one or both events happen is 1 - 5/12 = 7/12.

[winniethepooh]

Thanks Ian!

[edvhou812]

Probability Event A or B will happen is A+B - A*B.

Dice is 3: 1/6
Coin is Heads: 1/2

1/6+1/2 - 1/6*1/2

8/12-1/12 = 7/12

[Ashley@VeritasPrep]

Diagram as promised! Note that this applies to the P(A) + P(B) - P(A-and-B) method, not to the alternative (also good!) method Ian proposes (1 - P(neither).