Basic Statistics Question - Independent Events

[Alley_Dweller]

@Una_Persson

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.

Think of the case where the probabilities are higher. Instead of 5% and 10%, think 50% and 60%. It should become clearer.

[Thudlow_Boink]

@Una_Persson

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.

Probability can be counterintuitive, but it might help your intuition to reframe things in terms of frequencies (and whole numbers). Instead of saying event A has a 5% chance of occuring and event B a 10% chance, you could think that, out of every 100 days, there will be 5 days on which event A happens, and there will be 10 days on which event B happens. If you count up how many days there are out of the 100 on which either A or B happens, there’s no way it can be more than 15 days. (It might be less than 15 if there’s some overlap between the days when A happens and the days when B happens.)

[Una_Persson]

@Andy_L

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).

I did, and I thank you for the illustration, but it still just isn’t sticking with me.

Damn it, I’ve had math through partial differential equations and integral transformations, and some things still flummox me. Then again, it’s been nearly two decades since poor me was last in a math class (sobs…I’m so old…).

Ultrafilter and Alley Dweller, you’re correct, and I actually did Alley Dweller’s thought experiment, which is why I knew that I was wrong, and I needed to ask on here.

[Una_Persson]

And I really do appreciate the help you all have given me. Thank you very much. Even if I don’t develop the intuition I’d like on this specific problem, I do at least understand the solution.

[Little_Nemo]

14.5%.

4.5% chance of A alone occurring
9.5% chance of B alone occurring
0.5% chance of A and B occurring

[Little_Nemo]

And now the incredibly half-assed way I figured out those numbers.

I knew there was a 5% chance of A and a 10% chance of B based on the numbers given to me. And I knew there would be an overlap because some A results would also be B.

So I figured if 10% of the overall results would be B, then 10% of the A results would be B. Ten percent of the A results is .5%. So I knew the 5% of the results that were A were 4.5% A alone and .5% A and B.

And knowing that 10% of the overall results were B and .5% of the results were A and B, I knew that 9.5% of the overall results were B alone.

So I just added up my percentages:

4.5% chance of A alone occurring
9.5% chance of B alone occurring
0.5% chance of A and B occurring
14.5% total A, B, or A and B

The sound you’re now hearing is mathematicians weeping at this post.

[Thudlow_Boink]

@Little_Nemo

The sound you’re now hearing is mathematicians weeping at this post.

No they’re not. That’s a perfectly good, legitimate explanation.

[Little_Nemo]

@Una_Persson

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.

The percentages can’t add up to a total greater than the sum of the individual percentages.

Look at it this way. Imagine you have a jar full of a hundred coins. And you know that ten of the coins are Canadian. And you know that five of the coins are quarters.

If you took out all of the Canadian coins and all of the quarters in the jar, you obviously will not take out more than fifteen coins. (And you might take out less than fifteen because some of the coins might be Canadian quarters.)

[Little_Nemo]

@Thudlow_Boink

No they’re not. That’s a perfectly good, legitimate explanation.

I feel there is probably a regular procedure for figuring out the probabilities that is more elegant than the approach I used.

[Chronos]

I feel there is probably a regular procedure for figuring out the probabilities that is more elegant than the approach I used.

“Elegant” is a matter of taste. It’s a valid method, and in the end, that’s what matters.

In point of fact, there are two methods which are commonly used for this problem, and one of them is pretty much derived by doing what you just did, except in the general case so you can plug the numbers in at the end instead of at the beginning.

[cerberus]

linky

Event A is in yellow/orange B is in red/orange. AandB is in orange.

Pr{A}=.05: this includes both the AB intersect and the parts of A not overlapping B.
Pr{B}=.10: this includes both the AB intersect and the parts of B not overlapping A.

Part of the .05 is Pr{AB} and part of the .10 is Pr{AB}. Under independence,
Pr{AB}=.05*.10=.005.

So Pr{“A or B”} = Pr{A}+Pr{B without AB intersect} =
.05+(.10-.005) = .05+.995=.145.

And Pr{“A or B”}=Pr{A without AB intersect}+Pr{B}=
(.05-.005)+.10 = .145.

[nivlac]

@Little_Nemo

I feel there is probably a regular procedure for figuring out the probabilities that is more elegant than the approach I used.

What you did was fine. I didn’t weep. The more standard way (I wouldn’t say necessarily more elegant way) is to use the rules involving probabilities that are usually taught in a beginning stat or finite math course. Off the bat I can think of at least three ways to solve for P(A or B) given the conditions in the OP. I’m glad the correct solution was given after the bad answers in the first two posts.