Question

Seven unbiased coins are tossed $128$ times. In how many throws would you find at least three heads?

• A. $99$
• B. $102$
• C. $103$
• D. $104$

Answer 1

Answer: (A)

$99$

Solution:

Given, $p=q=\frac{1}{2},n=7,N=128$

$r=3,4,5,6,7$

$\therefore P(X\ge 0)=128[C_3^7{p}^3{q}^4+C_4^7{p}^4{q}^3+C_5^7{p}^5{q}^2+C_6^7{p}^{6}q+C_7^7{p}^7{q}^0]$

$=128[35{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^4+35{\left(\frac{1}{2}\right)}^4{\left(\frac{1}{2}\right)}^3+21{\left(\frac{1}{2}\right)}^5{\left(\frac{1}{2}\right)}^2+7{\left(\frac{1}{2}\right)}^6\left(\frac{1}{2}\right)+{\left(\frac{1}{2}\right)}^7]$

$=128\left[{\left(\frac{1}{2}\right)}^7\right(35+35+21+7+1\left)\right]$

$=(\frac{1}{128}\times 199)\times 128$

$=99$

Hence, A is the correct option.