The question involves 2 urns, URN1 and URN2. There are 18 balls of 4 variations in URN1: 3 Red Balls, 5 Orange Balls, 5 Yellow Balls and 5 Green Balls. There is 1 ball of unknown color in URN2, i.e. it can be any of the four colors. Mary dies if she picks a Orange ball and Sam dies if he picks a Red ball.
Mary picks 1 ball from URN1 and does not replace it. However, she takes the ball from URN2 and places it into URN1. Subsequently, Sam picks 1 ball from URN1. Given that Mary dies, what is the probability that Sam dies as well?
My logic was that the probability of Mary dying, picking a Orange ball, will not affect the probability of Sam dying, picking a Red Ball. Hence the answer should be: (3/18) + (1/18)*(1/4) = 13/72
So (probability of drawing a Red ball) + (Probability of drawing the random ball)*(Probability of random ball being red).
But I have been told that this answer was wrong and I am not sure why as well.
Edit: To clarify, Mary is able to transfer the ball from URN2 to URN 1 even if she draws the Orange Ball(i.e. Dies). Maybe she doesn't look at it just yet??
Your Math is correct, but your analysis is wrong. The events of Mary and Sam dying are not independent. The reason is that if Mary dies, then you know for sure that she took an Orange ball. This means that you are guaranteed that she did not take a Red ball. This is relevant to Sam's survival chances.
Since Mary died, (presumably) after the Urn-2 ball is placed in Urn-1, and since the presumption is that the Urn-2 ball has (in effect) a $(1/4)$ probability of being any one of the $4$ colors, what you have, right before Sam chooses a ball is that there are now $(3 + 1/4)$ balls out of $(18)$ that are Red.
So, given that Mary died, she did not take away a Red ball. Therefore, Sam's probability of dying is
$$\frac{3 + (1/4)}{18} = \frac{13}{72}.$$
Just to drive the point home, consider what Sam's probability of dying would be if Mary had lived. This would mean that she did not take an Orange ball. This would imply that she selected a Red ball, with probability $(3/13)$, and a non-Red ball with probability $(10/13)$.
This means that the expected number of Red balls after she does not die are $3 - (3/13) + (1/4) = \frac{156 - 12 + 13}{52} = \frac{157}{52}.$
Therefore, if Mary lives, Sam's probability of dying has changed to
$$\frac{157}{52 \times 18}.$$