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There are two coins. One is two headed coin, the other one is a fair coin. A coin is selected randomly and flipped.

a) what is the probability of selecting a fair coin?

b) what is the probability of head from the flipped coin?

c) what is the probability of selecting a fair coin and having head on the top?

So A and B are relatively simple, because A is $1/2$ and for B its $3/4$ but I am concerned with C

Shouldn't be the chance of having a fair coin $+$ head is basically $1/4$? Correct me if I am wrong but

If I want to pick a fair coin, the probability is $0.5$, lets say we picked it, and now we are going to throw it, isn't also chance between head and tail is $0.5$? Which effectively means that getting a head from a fair coin

the intersection between them is $0.5\times0.5$ which equals $0.25$?

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  • $\begingroup$ Is the answer to the c part, 1/3? $\endgroup$
    – Arsenic
    Jul 30, 2022 at 14:53
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    $\begingroup$ @Arsenic $P(A\land B)=P(A|B)P(B)$ so the probability of getting the fair coin and getting a head is the probability of getting a head, given that we have a fair coin ($\frac{1}{2}$), times the probability of getting a fair coin which is also $\frac{1}{2}$ so we get $\frac{1}{4}$. If we chose a coin, flipped it and got a head then the probability that the head came from the fair coin is $\frac{1}{3}$. $\endgroup$
    – John Douma
    Jul 30, 2022 at 15:02
  • $\begingroup$ @Arsenic The prof didn't give us any answer or key sol I am sorry $\endgroup$
    – Eimon
    Jul 30, 2022 at 15:02
  • $\begingroup$ @JohnDouma Yep! I presumed that the answer OP posted in the question was marked incorrect, so could only see conditional probability being a reason- which would have meant that the question wasn't framed correctly. $\endgroup$
    – Arsenic
    Jul 30, 2022 at 15:07

1 Answer 1

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Let $F=$ "select a fair coin", and $H=$ "appear head on top"

$$P(F\cap H)=P(H|F)\cdot P(F)=\frac{1}2\cdot\frac{1}2=\frac{1}4$$

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  • $\begingroup$ Thank you so much! $\endgroup$
    – Eimon
    Jul 30, 2022 at 15:11
  • $\begingroup$ You are most welcome! If the answer helps, please $\checkmark$ my answer :) @Eimon $\endgroup$
    – MathFail
    Jul 30, 2022 at 15:13
  • $\begingroup$ Done ^ Thanks again $\endgroup$
    – Eimon
    Jul 30, 2022 at 15:14
  • $\begingroup$ You are welcome! $\endgroup$
    – MathFail
    Jul 30, 2022 at 15:17
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    $\begingroup$ It may help to consider that it doesn't matter what the unfair coin is. It could be a two-tailed coin or a teddy bear; you are selecting (1/2) and then flipping (1/2). $\endgroup$
    – fectin
    Jul 30, 2022 at 23:13

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