The Student Room Group

Probability question

Following a flood 120 tins were recovered from Dharmesh's corner shop. Of the tins 50 contained pet food, 20 contained peas, 35 contained beans and 15 contained soup.

a) Dharmesh selects three tins at random (without replacement). Find the probability that one contains pet food, one contains peas and one contains beans. (answer = 0.125)

b) Find the probability that Dharmesh will have to open more than two tins before she finds one which does not contain pet food. (answer = 0.172)


They seem easy but somehow I keep on getting the wrong answers :s-smilie: My Stats exam is tomorrow so I'm really starting to panic!

Reply 1

first one:

50/120 * 20/119 * 35/118 * 3!

(the 3! on the end is because there are 3! ways of ordering that combination)


second one: 50/120 * 49/119 because this is the probability that the first two will be pet food, irrespective of whatever follows.

Reply 2

Well, you have the answers so perhaps we can tell where you're going wrong if you post your working or reasoning?

Reply 3

Dirac Delta Function
first one:

50/120 * 20/119 * 35/118 * 3!

(the 3! on the end is because there are 3! ways of ordering that combination)


second one: 50/120 * 49/119 because this is the probability that the first two will be pet food, irrespective of whatever follows.


Oooh, I now see the silly mistakes I had made :p: Thanks!

Reply 4

a) Write down all the different ways the three could come out:

Key: Ps= Peas P=Pet B=Beans

Ps P B
Ps B P
B Ps P
B P Ps
P Ps B
P B Ps Total 6 Outcomes

Then calculate the probabilty for each one by multiplying then adding all the sums. e.g. ps P B= 20/120 X 50/119 X 35/118 = 0.0208

Now if you think logically then the each branch should have the same answer

and because there are 6 branches you multiply 0.0208 by 6 which is 0.125

Reply 5

Thanks :smile:

I have another question now though:

D and C go to the cinema. Probability that D buys popcorn is 0.4. Probability that C buys popcorn is 0.7 if D buys and 0.35 if D doesn't. They are joined by S. Probability that S buys popcorn is 0.55 if both D and C buy popcorn, and 0.25 when exactly one of D and C buys popcorn.

Find the probability that C and S buys popcorn but D doesn't.

I tried doing this and did 0.6 x 0.35 x 0.25 = 0.0525, but apparently the answer is 0.105....?