1
$\begingroup$

We throw a coin 8 times. What is the probability of getting the same number of heads and tails, if on the last three tosses of a coin we got tails?

$\endgroup$
3
$\begingroup$

Getting same number of heads and tails means getting $4$ heads and $4$ tails. You already got $3$ tails. So, the problem is reduced to getting $4$ heads and $1$ tail in the first five tosses. That is nothing but $5C_4$$*$$1\over 2^5$ $= $$5\over 32$

$\endgroup$
1
$\begingroup$

Let $x = $ the number of tails that remain, and $y = $ the total number of heads, then: $3+x = y, 3+x+y = 8$. Thus $ x = 1, y = 4$, Thus the remaining one tail can occur in any of the first $5$ tosses. So the probability is: $5\cdot \left(\dfrac{1}{2}\right)^5$

$\endgroup$

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.