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In "Conditional probability with Bayes" theorem in Khan's academy, in 2nd experiment, where author has 2 fair coins, and 1 biased coin, he tries to calculate probability of biased coin, after first test: Heads.

Note: A biased coin is one which gives heads 2/3 times, and tail 1/3 times

p(B) = probability of biased coin
p(B/H) = probability of biased coin, given its Heads.
p(H) = probability of coin being Heads
p(H/B) = probability of coin being Heads, given its biased
p(B and H) = probability of coin being both biased and heads.

Event 1: Pick: p(B) = 1/3 (Since he has 2 fair coins and 1 biased coin) Event 2: Flip: Heads: p(H/B) = 2/3 (Since a biased coin gives heads 2/3 of the time) Now, what is p(B/H), that is probability being biased coin, given heads was outcome.

As per my understanding,
p(B and H) = p(H and B)

Thus
p(H and B) = p(H/B)p(B) = (2/3)(1/3) = 2/9
p(B and H) = p(B/H)p(H)

Now how do we calculate p(H)? How does p(B/H) become 4/10?

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  • $\begingroup$ I think you can just sum all the probabilities when you get H. p(H) = 1/2*2/3+1/3*2/3 $\endgroup$ – Michael Paris Mar 7 '18 at 15:48
  • $\begingroup$ Few people want to watch a 5 min video to find out what your question is. I would just cut the link and explain concisely what the problem is. $\endgroup$ – almagest Mar 7 '18 at 15:49
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You sum all the probabilities that result in head (H) \begin{equation} p(H) = 1/2 * 2/3 + 2/3 *1/3 = 5/9 \end{equation}

And so $p(B|H) = 2/9 * 9/5 = 2/5$

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  • $\begingroup$ your answer was spot on, and helped me understand bigger picture now. I have created an in depth article here. Please check and share your feedback if time permits. Thank you for your timely help. $\endgroup$ – Parthiban Rajendran Mar 8 '18 at 16:55

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