The question is broken down in three parts, I know the answers to all three questions, however can only work out (ii). It would be much appreciated if someone can break down the working out for (i) and (iii).
A student leaves her phone behind with probability of 1/4 each time she attends class. She sets out with her phone to attend 5 different classes.
(i) If she arrives home without her phone (after attending 5 classes), what is the probability that she left it in the 5th class? Answer: 0.104
(ii) What is the probability that she will leave her phone in the 5th class? Answer: 0.0791
(iii) If she arrives home without her phone and she is sure she had the phone after leaving the first class, what is the probability that she left it in the 5th class? Answer: 0.154
Ok, you have solved (ii) and need help for (i) and (iii)
In (i), it is known that she has left her phone, so what is being asked is
P[left phone in class 5|left phone in some class]
$= \frac{P( left\; phone\; in\; class\; 5)}{P(left\; phone\; in\; some\; class)} \quad = \dfrac {(3/4)^4(1/4)}{1-(3/4)^5}=\frac{81}{781}\; \approx 0.104$
For (iii), do it like (i) except that now there are only $4$ classes to consider instead of five.