Question

An unbiased coin is tossed $10$ times. By using binomial distribution, find the probability of getting at least $3$ heads.

• A. $\frac{110}{128}$.
• B. $\frac{51}{63}$.
• C. $\frac{121}{128}$.
• D. $\frac{150}{176}$.

Answer 1

The correct option is C $\frac{121}{128}$.

= 1 - { Pr(no head) + Pr(1 head ) + Pr(2 heads)

$=1-C(10,0)(1/2{)}^{10}+C(10,1\left)\right(1/2{)}^{10}+C(10,2)(1/2{)}^{10}$

$=1-(1/2{)}^{10}+10(1/2{)}^{10}+45(1/2{)}^{10}$

$=1-56/(2{)}^{10}=1-\left(56/1024\right)=968/1024$

$=\frac{121}{128}$