# Probability of exactly one student failing?

I have the following problem:

If the probability that student A will fail a certain statis- tics examination is 0.5, the probability that student B will fail the examination is 0.2, and the probability that both student A and student B will fail the examination is 0.1, what is the probability and that exactly one of the two students will fail the examina- tion?

I came up with the following solution: $$P(A) = 0.5; P(B) = 0.2$$ 'exactly one' means either A only fails or B only fails.

Event $$X_1$$: 'A only fails'$$P(X_1) = P(A) *P(B)^c = 0.5*0.8 = 0.4$$ Event $$X_2$$: 'B only fails' $$P(X_2) = P(A)^c * P(B) = 0.5*0.2 = 0.1$$ And therefore: $$P(X_1\lor X_2) = P(X_1)+P(X_2)-P(X_1\land X_2) = 0.4+0.1-0 = 0.5$$ My thoughts behind $$P(X_1 \land X_2) = 0$$ were that it is not possible that both only happens at the same time. I'm not fully sure whether that is correct. Should these logical thought be correct is the value of $$0.5$$ correct?

• Welcome to math.SE! Asking for correctness here is OK (and it is encouraged to provide your own ideas for every question, what did you try and what didn't work), but it would also be good in the future to write your solution using MathJax, instead of uploading pictures. – Nutle Nov 21 '18 at 15:54
• It'd be better to write your solution in the body of the question rather than linking to an image. (Also, the image should be rotated to make it easier to read.) – littleO Nov 21 '18 at 15:54
• I see. Thank you both, I edited my question now. – thebilly Nov 21 '18 at 16:10

You need to find $$P(A, \neg B) + P(\neg A, B)$$. We know $$P(A)$$, $$P(B)$$ and $$P(A, B)$$.
We also know that $$P(A,B)+P(A,\neg B)=P(A)$$. You can get the value $$P(A, \neg B)$$ from here. Likewise, we also know that $$P(A, B)+P(\neg A, B)=P(B)$$ and thus you can also get the value $$P(\neg A, B)$$.
Your $$P(X_1)=0.4$$ and $$P(X_2)=0.1$$ are right. But then it is asked for $$P(X_2\cap \overline X_1)+P(X_1\cap \overline X_2)$$
This is $$[P(X_1)-P(X_1\cap X_2)]+[P(X_2)-P(X_1\cap X_2)]=P(X_1)+P(X_2)-2\cdot P(X_1\cap X_2)$$
$$=0.2+0.5-2\cdot 0.1=0.5$$