joshjoshjosh

New member
Hi guys. I was stuck on this question and would like to get some help from the forum! The question is:

You ask your neighbour to water 3 sickly plants in your front yard while you are on vacation. Each of the plants will be alive with probability 0.9 if watered; it will be alive with probability 0.3 if not watered. You are 95 per cent certain that your neighbour will remember to water your plants.

Let X be the number of sickly plants that you will find to be alive when you return from your vacation.

(a) What is the probability P(X=2)
(b) If only one plant is found to be alive when you return, what is the conditional probability that your neighbour remembered to water the plants?

mmm4444bot

Super Moderator
Staff member
Hello joshjoshjosh. Are ya joshin'? What have you tried or thought about so far?

Please check out the forum's submission guidelines. There's also a link at the top of the front page (after you logged in), unless you closed it. Thank you! Jomo

Elite Member
Hi, welcome to the forum.
In order for you to receive good help we need to know where you are stuck. Please respond with the work you have done so far even if you know it's wrong -- that way we can start with what you need. Thanks

joshjoshjosh

New member
Hello @mmm4444bot and @Jomo,
I have tried the question so far.

For question (a)
Let A = you neighbour remembers, so P(A) = 0.95 and P(A') = 0.05.
Let B = your plant survive, so P(B|A) = 0.9 and P(B|A') = 0.3.
First of all, we need to find the probability that each of the plants will be alive. Using the theorem of total probability we have:
P(B) = P(A)P(B|A) + P(A')P(B|A') = 0.95 x 0.9 + 0.05 x 0.3 = 0.87
Then for 2 of the three plants to survive, this is the binomial probability:
P(X = 2) = (3 choose 2) x 0.87^2 x (1 - 0.87) = 0.295

For question (b) I'm not sure how to do it!

HallsofIvy

Elite Member
Let "p" be the probability your neighbor waters the plants and imagine doing this 1000 times. 1000p times he waters the plants, 1000(1- p) times he does not. The probability any one plant survives is (1000p(0.9)+ 1000(1- p)(0.3))/1000= (900p+ 300- 300p)/1000= 0.6p+ 0.3. The probability any one plant does not survive is 1- (0.6p+ 0.3)= 0.7- 0.6p.

The probability exactly two of the three plants survives is 3(0.6p+ 0.3)^2(0.7- 0.6p). Since that is what happened, set it equal to 1 and solve for p.

The probability exactly one of the three plants survives is 3(0.6p+ 0.3)(0.7- 0.6p)^2. Set that equal to 1 and solve