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probability of picking a fair coin= 2/4=1/2= probability of picking unfair coin

suppose the coin picked is fair one;

(1/2)*(1/2)^4= 0.5^5

coin picked is unfair one;

(1/2)*(0.6)^2*(0.4)^2

required probability = 0.03125+0.0288= 0.06005

I did exactly the same thing whilst calculating the probability, except for one thing:-
In either case(fair of unfair coin), two tails can turn up in 4 flips in exactly 6 ways (HHTT, HTHT, TTHH, THTH, THHT and HTTH) so the total probability will be 6 times the above value!!

what say??

i thought about it but cant reach a consensus logically so waiting for some comment describing the issue at hand.

I think Xlogic is correct , the result has to be multiplied by 6 as they are six favorable sets. Killer1387 has obtained the answer without considering the number of ways it can be achieved in 4 tosses. or in other words he has got the result for a single set like getting only a HHTT combination. So the final answer is definitely 0.06005*6

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I did similar to Neo.

P(fair) = (1/2)^4 * 4C2
= (1/2)^4 * 4!/2!2!
= (1/2)^4 * 6

P(unfair) = (2/5)^2 * (3/5)^2 * 6
= (6/25)^2 * 6

P(unfair or fair) = (6/25)^2 * 6 + (1/2)^4 *6
= 6*[(6/25)^2 + (1/2)^4)]

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I dont see as to why we need to consider arrangements here , we just have 2 tails out of 4.

btw whats the OA??

@Killer, you don't just have 2 tails out of 4. You have X number of ways of getting 2 out of 4.

If we simplified the question to: If I tossed a fair coin 3 times, what is the probability of heads showing up exactly twice?

The question is how many ways is this possible:
HHT or
HTH or
THH

Therefore we get: (1/2 * 1/2 * 1/2) 3 times
3*(1/2)^3

If the question was: A fair coin is tossed three times, what is the probability that only the first 2 tosses end up with heads.

Then this is more restrictive: HHT = 1/2 * 1/2 * 1/2 = (1/2)^3

Thoughts??

OA: None

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You have beautifully put it into words! Great Going!