1
$\begingroup$

Two coins are in a hat. The coins look alike, but one coin is fair (with probability 1/2 of Heads), while the other coin is biased, with probability 1/4 of Heads. One of the coins is randomly pulled from the hat, without knowing which of the two it is. Call the chosen coin “Coin A”.

Find the probability that in 10 flips of Coin A, there will be exactly 3 Heads.

PROPOSED SOLUTION:

P(H) = P(H|Fair)*P(Fair) + P(H|Biased)*P(Biased) = 1/2*1/2 + 1/8*1/2 = 3/8

P(T) = 1 - P(H) = 1 - 3/8 = 5/8

Now, P(3H in 10 flips) = (10C3)((3/8)^3)((5/8)^7) = 0.236

However the answer is given as 0.184.

Can anybody help understand the error in above approach. Thanks

$\endgroup$
2
$\begingroup$

You calculated the probability for the experiment in which ten times a coin is drawn from the hat, tossed and replaced. The question is about the experiment in which a single coin is drawn from the hat and tossed ten times.

If the difference is still not clear, first consider the easier case of $2$ tosses instead of $10$.

$\endgroup$
  • 1
    $\begingroup$ Thanks very much joriki. Understood the difference. $\endgroup$ – Rahul Aug 2 '15 at 13:01
1
$\begingroup$

$P( \texttt{3 Heads}) = P(\texttt{3 Heads|fair})P(\texttt{fair}) + P(\texttt{3 Heads|biased})P(\texttt{biased})$

$=1/2\cdot{10\choose 3}\cdot(1/2)^3\cdot(1/2)^7 + 1/2\cdot{10\choose 3}\cdot (1/4)^3\cdot (3/4)^7$

$=0.184$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.