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Question

Suppose you have two coins A and B the probability of head in A is $\frac{1}{4}$ and the probability of head in B is $\frac{3}{4}$. Now, suppose you have chosen a coin and tossed it two times. The output was head and head. What is the probability that you chose the coin B.

My Approach

I used Bayes' theorem ,

Let Event, $\text{output to be Head Head}=E$

Req'd probability=$P({B}\mid{E})$

$$P({B}\mid{E})=\frac{P({E}\mid{B})\times P(B)}{P({E}\mid{B})\times P(B)+P({E}\mid{A})\times P(A)}$$ $$=\frac{\frac{3}{4} \times \frac{3}{4} \times \frac{1}{2}} {{\frac{3}{4} \times \frac{3}{4} \times \frac{1}{2}}+{\frac{1}{4} \times \frac{1}{4} \times \frac{1}{2}}}$$

$$=\frac{9}{10}$$

Am i correct?

Community
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laura
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    The self-learning tag states that the tag should be used when the question is about the process of studying mathematics on your own. As for your question, you solved the problem correctly. – N. F. Taussig May 16 '18 at 21:25
  • i tagged it self learning because i did not know the answer and also unaware that my answer is correct or not.Can you suggest in which case should i tag self learning ...thanks – laura May 17 '18 at 05:19

2 Answers2

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I am not aware of Bayes' theorem but I have also reached the answer of $\frac{9}{10}$.

If you are using coin A, then the probability of getting double heads is $\frac{1}{4}*\frac{1}{4}=\frac{1}{16}$.

If you are using coin B, then the probability of getting double heads is $\frac{3}{4}*\frac{3}{4}=\frac{9}{16}$.

We can now take this as a ratio of probabilities:

$$\frac{1}{16}:\frac{9}{16}$$ $$=1:9$$ $\therefore$ the probability of the coin you chose being coin B is $\frac{9}{10}$.

(I don't know if this is a mathematically viable way of getting the answer. It is just something that came into my head so if this doesn't work then just mention it in the comments and I will gladly delete the post.)

Dan
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  • Well, it may not be formally correct, but it is a good grasp of the intuition behind the concepts. – Graham Kemp May 16 '18 at 21:29
  • That's more or less discovering Baye's theory on your own. The idea is the probability of $x$ given that $y$ happened is the of $y$ if $x$ happened divided prob of $y$ at all. Which is common sense. The probability of picking coin B and getting 2 H is $\frac 12\frac 34\frac 34 = \frac 9{32}$ The probability of 2H is prob of A2H + prob of B2H. Which is $\frac 12\frac 14*\frac 14 + \frac 9{32}=\frac{10}{32}$. So although there are 32 expected outcomes, only 10 of them involve 2H. And $9$ out of $10$ include B. So P=9/10 = $\frac {9/32}{10/32}$. – fleablood May 16 '18 at 22:45
  • +1. This's nice. It provides the problem intuition without to many symbols. Great. – Felix Marin May 17 '18 at 02:33
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Am i correct?

Yes, that is correct.

The answer is $\mathsf P(B\mid E)=9/10$.   Obtaining two heads is strong evidence that the coin is biased towards heads, so you should anticipate the answer will be somewhat greater than $\mathsf P(B)$.

By Bayes' Rule: $~\mathsf P(B\mid E) = \mathsf P(E\mid B)\cdot \mathsf P(B)~/~\mathsf P(E)$

By Law of Total Probability (and that events $A,B$ are disjoint and exhaustive (ie partition the space)): $~\mathsf P(E)=\mathsf P(E\cap B)+\mathsf P(E\cap A)$

So, putting this together: $$\mathsf P(B\mid E)=\dfrac{\mathsf P(E\mid B)~\mathsf P(B)}{\mathsf P(E\mid B)~\mathsf P(B)+\mathsf P(E\mid A)~\mathsf P(A)}$$

Everything else is just substituting the appropriate evaluations and doing the calculations, which you have done.

Graham Kemp
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    The question was "am I correct". So far as I can tell there are two possible answers. Either "Yes, you are" or "No, you are not". So I think this most certainly does provide an answer. – fleablood May 16 '18 at 22:27