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Statistics S1 Help needed (Jan 04) - "...given that..."

Good morning :smile:

The question I have is (in full):
John is absent minded. He also carries an umbrella but leaves it behind in a shop with probability 1/3. He leaves home with the umbrella and visits each of 3 shops.
a) Find the probability he leaves his umbrella in the second shop [3]
b) Find the probability he leaves his umbrella in one of the shops [3]
c) Given that he leaves his umbrella in one shop, find the probability that he leaves it in the second one [3]

I know S1 is simple, but part C of this question is the part I always seem to have a bit of trouble with. I just need to clear it up so that I know -exactly- what I need to do.

Many thanks

Reply 1

Jonny W

Define two events:

A = {John leaves the umbrella in second shop}
B = {John leaves the umbrella in one of the shops}

P(A | B)
= P(A and B)/P(B)
= P(A)/P(B)
= (2/3)(1/3) / [(1/3) + (2/3)(1/3) + (2/3)(2/3)(1/3)]
= 6/19

Reply 2

arktos

Sorry I'm really bad at S1, but If he left his umbrella in the first shop, he would have no umbrella to leave in the second shop wouldn't he?

Reply 3

Craver

Ohh, I was making it more complicated that it actually was. I assumed that because it was saying "leaves it in one shop" that he could leave it in 2 or even all 3 (by going back and collecting it first of all again obviously)... now it all makes sense.

Thanks!!