Need to find P(M|¬F)

Question

Jon is in a pit with 100 deadly scorpions, 60 of which are male and 40 of which are female. The male scorpion's bites are fatal 70% of the time and the females scorpions' bites are fatal 90% of the time. Jon escapes the pit, but is bitten once by one of the scorpions. Given Jon survives, what is the probability that the scorpion that but him was male? (Assume the scorpion that bit Jon was chosen uniformly at random from the 100 in the pit)

My answer:

I used a tree diagram in my working to solve out P(M) = 0.6, P(¬M) = 0.4, P(F|M) = 0.42, P(¬F|M) = 0.18, P(F|¬M) = 0.36, P(¬F|¬M) = 0.04 (I uses '¬' for 'not' since I couldn't find any better 'not' symbol).

M= Male scorpion

¬M= Not Male scorpion or Female scorpion

F= Fatal bite, dies

¬F= Not Fatal bite, Survives

I need to find P(M|¬F), So I used Bayes theorem found the result 0.87

Can anyone confirm if it is the correct result. Thanks!

Answer 1

Maybe I'm missing something, but here's how I understood the problem and solved it.

There's a 3/5 (.6) chance he was bitten by male. 2/5 (.4) chance he was bitten by female. There's a 5/5 (1.0) chance he was bitten at all.

We know he didn't die, and we know males aren't lethal with .3 probability, and females are nonlethal with .1 probability.

Consequently, the probability of him not dying is (.4*.1) + (.6*.3) = 11/50

Since we're wondering about what the probability it was a nonlethal bite from male scorpion, we take (9/50)/(11/50), where 9/50 is the probability of nonlethal male bite, and we get .81818181 repeating, or 9/11.

Answer 2

It looks like you mixed up $P(F \mid M)$ and $P(F \cap M)$. Defining $F$ to be "The bite was fatal" and $M$ to be "Jon was bitten by a male scorpion", we have:

• $P(M) = 0.6, P(\lnot M) = 0.4$. This is the distribution by gender of the scorpions.
• $P(F \mid M) = 0.7, P(\lnot F \mid M) = 0.3$. $70\%$ of male scorpion bites are lethal, thus $30\%$ are nonlethal.
• $P(F \mid \lnot M) = 0.9, P(\lnot F \mid \lnot M) = 0.1$. $90\%$ of female scorpion bites are lethal, thus $10\%$ are nonlethal.

We need to find $P(M \mid \lnot F)$, as you say. Using Bayes' theorem, we have $$ P(M \mid \lnot F) = \frac{P(\lnot F \mid M)P(M)}{P(\lnot F)} $$ We need to calculate $P(\lnot F)$. Using the law of total probability, we can do this with \begin{align*} P(\lnot F) &= P(\lnot F \mid M)P(M) + P(\lnot F \mid \lnot M)P(\lnot M) \\ &=(0.3)(0.6) + (0.1)(0.4) \\ &= 0.22 \end{align*} Putting the values into Bayes' theorem, we have \begin{align*} P(M \mid \lnot F) &= \frac{P(\lnot F \mid M)P(M)}{P(\lnot F)} \\ &= \frac{(0.3)(0.6)}{0.22} \\ &= \frac{0.18}{0.22} \\ &= \frac{9}{11} = 0.\overline{81} \end{align*} which is the final answer.