I would just like to double check that I have completed this question correctly. I am new to Baye's theorem and find the variables a bit confusing, particularly what a general rule is for determining A and B.
"Blofeld captures James Bond and places him in a pit with 100 deadly scorpions, 60 of which are male and 40 of which are female. The male scorpions’ bites are fatal 70% of the time and the female scorpions’ bites are fatal 90% of the time. Bond escapes the pit, but is bitten once by one of the scorpions. Given that Bond survives, what is the probability that the scorpion that bit him was male? (Assume the scorpion that bit Bond was chosen uniformly at random from the 100 in the pit.)"
So I did:
P(A1) = 60% (Scorpion is male)
P(A2) = 40% (Scorpion is female)
P(B|A1) = 30% (Bond survives given the scorpion was male)
P(B|A2) = 10% (Bond survives given the scorpion was female)
we want to find P(A1|B)
so
(.6)(.3)/[(.6)(.3)+(,4)(.10)
=.69
69% chance that the scorpion is male.
I would like to double check that this is correct
Baye's Rule: $P(A_1|B)=\frac{P(B|A_1)P(A_1)}{P(B|A_1)P(A_1)+P(B|A_2)P(A_2)}$
A1 denotes the event that the scorpion is male. $P(A_1)=0.6$
A2 denotes the event that the scorpion is female. $P(A_2)=0.4$
B denotes the event that the scorpion is non-fatal.
The probability that a male scorpion is non-fatal: $P(B|A_1)=0.3$
The probability that a female scorpion is non-fatal: $P(B|A_2)=0.1$
So, $P(A_1|B)=P(A_1|B)=P(A_1|B)=\frac{P(B|A_1)P(A_1)}{P(B|A_1)P(A_1)+P(B|A_2)P(A_2)} = \frac{0.3*0.6}{0.3*0.6+0.1*0.4}=\frac{9}{11}=0.818$
Therefore, there's a 81.8% chance that the scorpion is male given that it is non-fatal.