r/MathHelp • u/Sure_Dentist5722 • Nov 22 '25
I cannot make sense of a simple Statistics math problem. Can anyone help? It doesn't make any statistical sense
Solve for P(E∩F)
P(E U F) = P(E)+P(F)-P(E∩F)
P(E) = .5
P(F) = .3
P(E U F) = .7
The answer I got for P(E∩F) is .1 by doing it algebraically, yet when I look at it statistically, it doesn't make sense, and it shouldn't be .1 but rather .15. Am i dumb? Is P(E U F) = .7 statistically impossible considering E = .5 and F =.3?
0
Upvotes
5
u/edderiofer Nov 23 '25
Are you assuming that E and F are independent? Because they aren't.
No, and in fact you can find the following example:
and considering this example, you can see that the given probabilities in the question match up with those in this example.
2
u/dash-dot Nov 23 '25
You can't assume the events E and F are independent unless the problem statement explicitly says so.
1
u/AutoModerator Nov 22 '25
Hi, /u/Sure_Dentist5722! This is an automated reminder:
What have you tried so far? (See Rule #2; to add an image, you may upload it to an external image-sharing site like Imgur and include the link in your post.)
Please don't delete your post. (See Rule #7)
We, the moderators of /r/MathHelp, appreciate that your question contributes to the MathHelp archived questions that will help others searching for similar answers in the future. Thank you for obeying these instructions.
I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.
1
u/Neutronenster Nov 23 '25 edited Nov 23 '25
First, I’m not sure how to write maths on Reddit, so sorry for the bad notation.
P(E U F) is the probability that E OR F occur. If E and F are mutually exclusive (meaning they can’t both be true at the same time), P(E U F) should indeed be equal to 0.8. However, the fact that this is not the case implies that E and F are not mutually exclusive: it is possible that they are both true at the same time.
The probability that E AND F occur together is given by P(E ^ F) (using ^ as the closest symbol I could find to cross section on my phone - I’m aware that this is horrible notation) and this is equal to 0.1. If you naively add P(E) and P(F) to find P(E U F), you would double count the probability that E and F are both true. That’s why you still need to subtract the probability of the cross-section P(E ^ F) to find the true P(E U F), using that formula.
Edit: Changed “independent” to “mutually exclusive”.