Say there are two absolutely independent events with the following probabilities of occurring within any given day:
Event A: 5% chance.
Event B: 10% chance.
What is the probability that A, B, or A AND B will occur in that day? This is basic stuff but for some reason I keep getting non-logical answers.
Or in other words, 1.0 minus the probability that nothing happens at all?
Off the top of my head:
A OR B => Pa + Pb = 0.05 + 0.1 = 0.15 (15%)
A AND B => Pa * Pb = 0.05 + 0.1 = 0.005 (0.5%)
A OR B OR (A AND B) => 0.15 + 0.005 = 0.155 (15.5%)
15%, if they are mutually exclusive (e.g. assuming the events are something like “you win the lottery” and “there is an earthquake”)
http://cns2.uni.edu/~campbell/mdm/prob.html
We can ignore A AND B because it contains all of the events that A OR B contains, and we only care about the probability of ANY of the conditions happening.
Though I think I missed a case here… what it should really be is:
A OR B OR ((A AND B) AND (NOT A ONLY) AND (NOT B ONLY))
I disagree. “A AND B” includes “A OR B”, so I think you have some double counting there.
E.g. if, on a certain day, You Win The Lottery AND There Is An Earthquake, then it is necessarily the case that You Win The Lottery OR There Is An Earthquake.
To be clear, situations that are not mutually exclusive are things where either there is some overlap in conditions (e.g., the chance of drawing a black card out of a deck and the chance of drawing an ace, since there ARE black aces), or where there is a causal relationship between the two probabilities (for example, the chance of it raining heavily versus the chance of being in a traffic accident, where rain increases the chance of a traffic accident by some percentage, or the chance of having more than 5 children versus the chance of paying more than 100,000 dollars for the totality of all your childrens’ educations)
(At least one of A and B) is equivalent to (not ( (not A) and (not B) ) ) .
In other words, the probability of nothing happening is the probability of A not happening, and B not happening. An “and” on independent events is just a multiplication, so the probability of nothing happening is (1 - 0.1) * (1 - 0.05) = 0.9*0.95 = 0.855 . That’s the probability of nothing happening, so the probability of something happening is 0.145 , or 14.5% .
robert_columbia, you’re using a funny definition of “mutually exclusive” there. The way everyone else uses that term, it means that it’s impossible for both things to happen. “The baby will be a boy” and “the baby will have blue eyes” are not mutually exclusive. “The baby will be a boy” and “the baby will be a girl” are. Given that the OP has said that the events are independent, we know they’re not mutually exclusive.
This can be expressed as
A OR B OR ((A AND B) AND ~(A AND ~B) AND ~(B AND ~A))
…
distribute the NOT’s
…
A OR B OR ((A AND B) AND (~A OR B) AND (~B OR A))
…
do some Truth Tables (I did it by hand)
…
(A OR B) OR (A AND B)
…
do some more truth tables
…
A OR B
Thanks. I think my brain farted for a minute. I meant that there probabilities were completely independent.
I stand behind my calculations. The chance of something happening is the chance that the 5% event will happen, plus the chance that the 10% event will happen.
Let’s say there is a 60% chance Event A happens and a 70% chance Event B happens, does that mean there is a 130% chance that something happens?
There are two key facts here:
[ol]
[li]By definition, A and B are independent if P(A and B) = P(A)P(B).[/li][li]The inclusion-exclusion principle, which states that P(A or B) = P(A) + P(B) - P(A and B).[/li][/ol]
We have that P(A) = 0.1 and P(B) = 0.05, so P(A and B) = 0.005. Therefore, P(A or B) = 0.145.
Edit: Note that this agrees with what Chronos got. I think my method is a little more straightforward, but both are correct.
Not quite. Your figure for A or B double counts the situations where both happen, and I’m not quite sure what your logic is on the third line.
No. Just no.
Let’s say you are flipping a two-side coin and rolling a six-sided die.
The odds of flipping heads are 50%, the odds of rolling a 1 are 16.7% (rounded).
Using robert_columbia’s calculations, there is a 66.7% chance of getting heads or one or both.
Using chronos’s calculations, the odds are
1 - ((1-0.5) x (1-.167)) = 1 - (0.5 x .833) = 1 - .4165 = 0.5835
Let’s see. The possible outcomes are:
h,1 *
h,2 *
h,3 *
h,4 *
h,5 *
h,6 *
t,1 *
t,2
t,3
t,4
t,5
t,6
So there are 12 possible outcomes. I have placed an asterisk next to each outcome that satisfies our conditions ((heads or 1) or (heads and one)).
So 7 out of 12 possible outcomes satisfy our conditions.
7/12 = .583 = 58.3% (rounded).
Two ways to get the same answer:
Not A = .95; Not B = .90; (Not A) AND (Not B) = .95*.90 = .855; So A OR B OR Both = 1 - .855 = 0.145
The other way:
.05 + .10 - .005 = .145
(Add the probabilities of A and B; subtract the probability of A AND B which was just counted TWICE.)
Agree. One way to think about it is to imagine a 20 x 20 square matrix. There’s 10% chance of event A, so paint 2 rows of the square to represent that. There’s a 5% chance of event B, so paint 1 column of the square. What percentage of the matrix is painted - .05 + .1 - .005 or 14.5%.
The total probability for event A is .05 - the event AB is a subset of the event A, which is itself a subset of the universal set. So then
Pr{A} = Pr{ AandNotB } + Pr{ AandB } = .05 and
Pr{B} = Pr{ BandNotA } + PR{ AandB } = .10 .
So events A and B both contain a copy of AandB. So then
PR{A}+PR{B} conta
The total probability for event A is .05 - the event AB is a subset of the event A, which is itself a subset of the universal set. So then
Pr{A} = Pr{ AandNotB } + Pr{ AandB } = .05 and
Pr{B} = Pr{ BandNotA } + PR{ AandB } = .10 .
So events A and B both contain a copy of AandB. So then
PR{A}+PR{B} contains two copies of PR{AandB}, so then
PR{ “A or B” } = PR{A}+PR{B}-PR{AandB}
= .05+.10-PR{AandB}.
But under the mutual independendence of the event pair A,B,
PR{AandB}=PR{A}PR{B}=.05.10=.005 ,
so then
PR{“A or B”} = .05+.10-.005 = .145.
Thank you very much for the responses. Where I’m hung up is that I still have this “gut feel” that the total probability should be greater than the sum of the two. In other words, I was getting a total probability which was greater than 0.05+0.1. However, when I did the calcs I was getting 0.145.
I don’t understand why it’s not something > (0.1+0.05), other than of course it’s the way it is. It shows you why you can’t trust your gut in math sometimes.
Una, did you look at my geometric interpretation of the problem - I think that explains why simply adding the probabilities would be double-counting the overlap. That might help make the answer more intuitive (or it might not - there are lots of different ways to understand a math issue; if my suggestion doesn’t help you, then someone else will doubtless have another way to think about it that does help).
Imagine that you have a standard six-sided die where the numbers 1, 2, 3, and 4 have been painted red, and the numbers 5 and 6 have been painted blue. Let A be the event that you roll an odd number, and let B be the event that you roll a red number. P(A) = 1/2, P(B) = 2/3, and P(A and B) = 1/3, so A and B are independent. If the probability of P(A or B) were at least P(A) + P(B), then you’d have P(A or B) no smaller than 7/6, which is absurd.
What’s going wrong, as mentioned, is that there are some numbers which are red and odd. If you just count up the number of outcomes that lead to each event, you’ll double count these numbers, and so you need to correct your calculations by subtracting off the double counted part. So we have that P(A or B) = P(A) + P(B) - P(A and B). Does that make it clearer?