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2 votes
2 votes
Sue has 20 biscuits in a tin.

There are: 12 plain biscuits
5 chocolate biscuits
3 currant biscuits
Sue takes at random two biscuits from the tin.
Work out the probability that the two biscuits were not the same type.

User Paceaux
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1 Answer

20 votes
20 votes

Answer: 111/380

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Step-by-step explanation:

Define the following three events:

  • A = plain biscuit
  • B = chocolate biscuit
  • C = currant biscuit

Notation like P(A) = 12/20 means the probability of event A (aka picking the plain biscuit) is 12/20 since there are 12 plain biscuits out of 20 total. I'll leave the fraction un-reduced so we can see where the 12 and 20 come from. Going from 12/20 to 3/5 has us lose vital information about the original values.

Notation like AB means a plain biscuit was selected first and then a chocolate biscuit is selected second in that exact order. I'm assuming the first selection is not replaced. This makes the probability of the second selection dependent on the first.

So,

P(AB) = P(A)*P(B given A) = (12/20)*(5/19) = 3/19

P(AC) = P(A)*P(C given A) = (12/20)*(3/19) = 9/95

P(BC) = P(B)*P(C given B) = (5/20)*(3/19) = 3/76

The last step is to add those results to get the probability of event AB, event BC or event AC of occurring. Addition is possible since the events are mutually exclusive.

P(AB)+P(AC)+P(BC) = 3/19 + 9/95 + 3/76 = 111/380

I've skipped a few steps, so let me know if you need to see the finer details. Also feel free to ask about any questions in general if they come up.

User Nisarg Thakkar
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2.7k points