Question

One hundred identical coins each with probability $p$ of showing up heads are tossed once. If

$0<p<1$ and the probability of heads showing on $50$ coins is equal to that of heads showing on $51$ coins, then find the value of $p$.

Open in App

Solution

Verified by Toppr

We have $_{100}C_{50}p_{50}(1−p)_{50}=_{100}C_{51}p_{51}(1−p)_{49}$

or $p1−p =51!49!100! ×100!50!50! =5150 $

$⇒51−51p=50p$

$⇒51=50p+51p$

$⇒101p=51$

$∴p=10151 $

Was this answer helpful?

0

0