Conditional Probability of multiple independent events.

Question

This problem is from my teacher and I think their answer is wrong.

The problem is in the context of trapping rats.

I agree with my teacher for the answer in part A as that is just $$P(R=1)=3(0.3\cdot0.7\cdot0.7)=0.441$$

I do not agree with my teacher for the answer in part B. The answer my teacher gave me was equal to 0.63 using the probability tree below.

I found an alternate answer using this equality given to me.

$$P(X|Y)=\frac{P(X\cap{Y})}{P(Y)}$$

The question wants

$$P(\text{2 rats, given rat is caught in first trap})=\frac{{P(\text{2 rats }\cap{\text{ rat is caught in first trap}})}}{P(\text{rat is caught in first trap})}$$

So by using substitution

$P(\text{2 rats }\cap{\text{ rat is caught in first trap}})=2\cdot0.63=0.126$

$P(\text{rat is caught in first trap})=0.3$

$P(\text{2 rats, given rat is caught in first trap})=\frac{0.126}{0.3}=0.42$

I belive that my teacher incorrectly added the third possibility of 2 rats being caught which does not satisfy the fact that a rat was caught in the first trap. This leads to $P(\text{2 rats} \cap \text{rat is caught in first trap})=0.189$ instead of $P(\text{2 rats} \cap \text{rat is caught in first trap})=0.126$

After I asked about the answer to this question, my teacher checked the answer with another teacher and they insist that they are correct.

Am I missing something obvious and if so what? or is the teacher's answer incorrect?

Answer 1

You identified correctly the mistake your teacher made.

You can derive your result using the values of your teacher by correctly restricting the consideration to the upper branch of the depicted tree:

• $P(R=2) = 0.189$
• $\color{red}{P(R'RR) = 0.063}$ does not belong to the event $1st\, R$ $$P(R=2\, | \, 1st\, R)= \frac{(R=2) \cap (1st\, R)}{P(1st \,R)} = \frac{0.189 \color{red}{- 0.063}}{0.3} = 0.42$$