Hi
I have a 50% probability of getting award A and a 50% probability of getting award B.
What's the probablity of me getting either A or B?

Thanks

It is just 50%.. right?

100%. Is this a joke or did you misstate the question? You have described a coin flip.

Are the awards determined independently? Or do they each happen at a 50% probability on the same trial such that if one does not happen, then the other must happen?

1. If the Awards are NOT independent, such that you mean the question like flip a coin and if it is heads, then get Award A and if it is tails, then get Award B, then there is a 50% chance of getting each award and a 100% chance of getting either award.

2. If the awards are independent (like flip 2 different coins, one to determine each), then:

.....(a) The chance of getting neither award is 0.5 * 0.5 = 0.25 = 25%, so, therefore, the chance to get at least one award is 1 - 0.25 = 0.75 = 75%.

.....(b) The chance to win exactly one award is 50%:

............(i) 0.5 (win Award A) * 0.5 (lose Award B) = 0.25
................0.5 (win Award B) * 0.5 (lose Award A) = 0.25
................0.25 + 0.25 = 0.5 = 50%

...........OR

............(ii) Win Both Awards = 0.5 * 0.5 = 0.25 = 25%
.................Lose Both Awards = 0.5 * 0.5 = 0.25 = 25%
.................Win Exactly One Award = 1 - 0.25 - 0.25 = 0.5 = 50%

Assume the awards are not mutually exclusive (you can get both)

A: Guess right heads or tails with a penny toss
B. Guess right heads or tails with a nickel toss

Let a = P(A award) = 0.50, b = P(B award) = 0.50.

1. P(A only) = a AND not b = 0.5 * 0.5 = 0.25
2. P(B only) = not a AND b = 0.5 * 0.5 0.25
3. P(A and B) = a AND b = 0.5 * 0.5 = 0.25
4. P(neither A or B) = not a AND not b = 0.5 *0.5 = 0.25

Note: 1.+2.+3. +4.=1.0 (=all possibilities)

5. P(A or B but not both) = a only + b only = 1. + 2. = 0.50
6. P(A or B or both) = a only + b only + a AND b = .1.+2.+3 = 0.25 + 0.25+ 0.25 = 0.75

Edit: Didn't see Lego's response but I'll leave this up as it shows a slightly different way of showing all possibilities.

Not a joke. And not a coinflip

Lego05 and statmanhal.
Thanks for your replies.
Makes perfect sense

Sorry I just assumed you meant mutually exclusive.

If the probability of getting award A is 16%, 24% to get award B and probability of getting both A and B is 11%,
what is the probability of getting at least one award?


Thanks

These figures would mean they are not independent, so we need to know the relationship and the selection method. If they were independent then the probability of both is A * B, not 11%.

Well, can we just imagine that the probabilities of getting award A is 16%, 24% for B, but due to irrational psychological factors (for example), the probability of a jury granting both awards is 11%?

Sure, but that makes them not independent. So I don't think your question can be answered without knowing the relationship and how they affect each other. The cases are A not B, B not A, neither, and both. We don't know the first two of those (and I'm not sure we know the neither case).

Just use the inclusion-exclusion principle:

P(A OR B) = P(A) + P(B) - P(A AND B)

Actually, we know all the four quantities above and it should pretty obvious how to calculate them. If we win A 16% of the times and we win both A and B 11% of the times, we obviously win A but not B 5% of the times. Also without having knowledge of the inclusion-exclusion principle, it should be obvious that:

P(A OR B) = P(A NOT B) + P(B NOT A) + P(A AND B)

which reduces to the formula I gave above.

Hi
Thanks for you reply
A question:
Doesnt the above calculation remove the case where you get both A&B?
A&B should be included when the question is "At least" meaning= A or B or A&B

No, doesn't remove the case, since P(A) + P(B) double counts the time where you win both A and B.

However, imagine this equivalent problem. There are 100 people in a room. 16 of them have the object A. 24 have the object B. 11 have both objects.

How many have the object A, but don't have B?
How many have the object A, but don't have B?
How many have at least one of them?
How many have none of them?

These are trivial questions. However, let's see them.

- Since 16 people have A and 11 of them have both A and B, we are left with just 5 people with only A.
- At the same way, we have 24-11=13 people having B but not A
- Finally, we have 5 people with A but not B, 13 people with B but not A and 11 people with both of them -> 29 people with at least A or B.
- 71 people have nothing.

See where IEP comes from? When we add 15+24 we count the guys that have both A and B twice since they belong to both the 15 and the 24. So we have to subtract them to have the guys having at least one of them.

Thanks, I see now.

Nickthegeek
Ok so -P(A and B) just sets the degree to which A and B are overlapping (in a Venn diagram). It doesnt remove it.

Thanks

It doesn't "set the degree" of anything, it makes you not count the overlapping region more than once.

If you flip a coin twice, the chance of at least one Tails is not 100%, it's (50+50-25)%. The overlap is 25%, which gets counted twice if you don't subtract it. Alternatively, you don't need to subtract anything if you add each region separately (but it's often easier to over-add and then subtract).

Thanks

The easy way to do a problem like this is with a table:

This has what we know filled in, including that the totals have to add up to 1 in each direction. With the entire table filled in, we get:

Then pick the probability of interest, which is 1-[P(notA) AND P(notB)] = 1-.71 = .29

Loden Pants
Thanks for the tip.
I appreciate it.