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A very frustrated customer is trying to find an electronic receipt on their phone so they can return an item. The trouble is that the customer has three email accounts and can't remember which, one, account it was sent to. The customer assumes that there is an equal probability for each account. To add complexity the phone only has enough battery power to search one of the three email accounts

Unfortunately, the store says that this is the last day it will accept the return, and it is only 3 minutes till close, so there will not be any chance to get to a charger. The customer randomly decides to search one of the emails, without any bias.

Suppose, due to the organization and the various spam filters of the accounts, the chance of finding the receipt even if they were to search in the correct account is not a guarantee. The probability of finding it in account 1, assuming it was sent there is 62%, 54% if in account 2 and 56% for account 3.

Part (a) What is the probability that if the receipt is in account 2 that the customer will find it?

I think this one is (1/3)*(0.54) The third is for choosing the account and then times 0.54 is the prob for finding from the question. but the question seems like a conditional probability so I'm not sure.

Part (b) Calculate the probability the receipt was in account 2, if the search in account 2 is unsuccessful.

I think this one is ((1/3)(0.16))/((1/3)(0.31)+(1/3)(0.16)+(1/3)(0.49))

Part(c) What is the probability this person finds the email?

(1/3)(0.69)+(1/3)(0.84)+(1/3)(0.51) because we can either find it in account 1 or 2 or 3

Any help would be appreciated!

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    $\begingroup$ Please use MathJax to format math.meta.stackexchange.com/questions/5020/…. $\endgroup$
    – Shuri2060
    Jul 8 '17 at 2:35
  • $\begingroup$ I agree with your answers to (a) and (c), although I'll have to think about (b) $\endgroup$
    – Shuri2060
    Jul 8 '17 at 2:37
  • $\begingroup$ For account 1 is it 62% or 69%? And for account 2 is it 54% or 84%? ANd for account 3 is it 56% or 51%? $\endgroup$
    – Bram28
    Jul 8 '17 at 2:48
  • $\begingroup$ @Bram28 didn't spot that myself. Yes, good point. $\endgroup$
    – Shuri2060
    Jul 8 '17 at 2:49
  • $\begingroup$ I think I put the wrong numbers $\endgroup$ Jul 8 '17 at 2:51
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For b) you should get:

$$\frac{\frac{1}{3}\cdot 0.46}{\frac{1}{3}+\frac{1}{3}\cdot 0.46 +\frac{1}{3}}$$

The denominator is the chance of it not finding the receipt in account 2 when searching account 2. So if it is in account 1 or 3, you are certain to not find it when searching account 2.

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  • $\begingroup$ +1 - I agree with this. The denominator represents all of the cases where you can be unsuccessful. If you search in ac 2 while the receipt is in ac 1, then you are going to fail all the time, and likewise if it is in 3. $\endgroup$
    – Shuri2060
    Jul 8 '17 at 2:59

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