Question

If we take $1280$ sets each of $10$ tosses of a fair coin, the number of sets we expect to get $7$ heads and $3$ tails is

• A. $450$
• B. $300$
• C. $150$
• D. $75$

Answer 1

The correct option is C $150$

$P$(head with a fair coin) is $p=\frac{1}{2}$
$P$(tail with a fair coin) is $q=\frac{1}{2}$
Probability of $7$ heads and $3$ tails in $10$ tosses of a fair coin is equal to ${}^{10}{C}_7{\left(p\right)}^7{\left(q\right)}^3=\frac{10!}{(10-7)!7!}{\left(\frac{1}{2}\right)}^7{\left(\frac{1}{2}\right)}^3=\frac{120}{1024}$
Thus, out of $1280$ sets, each of $10$ tosses of a fair coin we expect $7$ heads and $3$ tails in $1280\times \left(\frac{120}{1024}\right)=150$ sets