I'm trying to solve the following probability problem:
A painter wishes to finish his two paintings on time for a client. He calculates that if he works hard on one of the paintings, he will have the probability of 0.8 to finish it, but only 0.4 of finishing the other one. If he tries to work hard on both of them, then he has a probability of 0.6 of finishing them each.
If he fails to deliver either of the paintings, he is given an extra chance to finish them by another date. If he fails to finish only one, he has the probability of 0.9 of finishing it on time. If he failed to finish both paintings, he will finish both of them with probability of 0.6.
Which method should he adopt to increase his chance of finishing both paintings at some time? [Assume work done on one painting is independent of the other.]
At the moment I'm trying to figure out from where I should start.
From reading this problem I managed to gather the following information:
Let A -> event of finishing painting A
Let B -> event of finishing painting B
So:
$P(B) = 0.8$
$P(A|B) = 0.4$
$P(A)+P(B) = 0.6$ {Not sure its correct as in the question it is worded as: "he has a probability of 0.6 of finishing them each"}
$P(B|A\cap B') = 0.9$
$P(A\cap B|A'\cap B') = 0.6$
From the above (if correct), I was thinking that in order to find the best option for the painter to finish both his paintings, I need to make use of the law of total probability. However, when I tried to use it I got stuck so I think I'm wrong in assuming this question is related to the law of total probability. Can someone guide me on how I shall start on finding the painter's best option. Thanks
No, I think you have to calculate separately
and compare.
Once he has chosen a strategy, you should interpret the events of finishing painting A and painting B to be independent with the given probabilities. So for the first strategy $P(A_1)=0.8$, $P(B_1)=0.4$, and they are independent so $P(A_1\cap B_1)=0.32$. Here I am writing $A_1$ for the event of finishing painting A by the first date, etc - your post has the same A for the first and second date, which is confusing.
Now $P(B_2\mid A_1\cap B_1')=0.9$ and $P(A_2\cap B_2\mid A_1'\cap B_1')=0.6$. But don't forget that also $P(A_2\mid A_1'\cap B_1)=0.9$ - even if he works hard on A, it is possible (by my interpretation of the problem) that he will finish B but not A by the first date, and then he still has a $0.9$ chance to finish A by the second date.
What you want is the total probability of finishing both by the second dat, which is given by $$P(A_1\cap B_1)+P(A_1\cap B_1'\cap B_2)+P(A_1'\cap B_1\cap A_2)+P(A_1'\cap B_1'\cap A_2\cap B_2),$$ since these cover all the ways that can happen.
Once you've calculated this, you now do the same for the other strategy. All that's different for the second strategy is that $P(A_1)=P(B_1)=0.6$.