1. I agree with 0.335..
Hello,
I have 2 questions. I have answered the one already but i am not quite sure though
1) A student is going to sit for an exam on Calculus. If he studies the night before the exam, he is got 0.99 probability to pass it. But if he choose to go to his friend's party instead, he is got 0.5 probability to pass it. In order to decide what to do , he tosses a coin. Finally , the next day he succesfully passes the exam. What is the probability that he gone to the party ?
My answer here is 0.335... If its wrong ill write down exactly what method i followed (i used cond. probability by the way for this) .
2) This is a bit tricky...
John and Mary, are going to sit for an exam. The possible results that they can get are A,B and C. The possibility that John gets a B is 0.3. The possibility that Mary gets a B is 0.4. The possibility that none of them gets an A, but at least one of them get a B is 0.1. What is the possibility that at least one of them gets a B, but none of them gets a C ? .
when i read it the first time, i thought, "oh well conditional probability..." but i dont think so.
This is my approach:
For John: P(John_B)=0.3
For Mary: P(Mary_B)=0.4
I also understood that:
[P(John_B) ∪ P(Mary_B)] ∩ P(A') = 0.1
And
We are looking for:
[P(John_B) ∪ P(Mary_B)] ∩ P(C') = ?
. So if my speculation is right, i should find P(A') and then P(C')... or something, but idk how. Any ideas there ?
Thanks !
On second question:
Let P(XY) be the probability that John gets X and Mary gets X.
We are given that P(BA)+P(BB)+P(BC)=0.3, P(AB)+P(BB)+P(CB)=0.4, P(BC)+P(BB)+P(CB)=0.1.
Finally P(BA)+P(BB)+P(AB)=(0.3+0.4−0.1)=0.6.
This looks better, what do you think people ?
My first approach looks quite silly to be honest...