Question

A fair coin is tossed 8 times, find the probability of at least six heads       

Answer 1

Let X denote the number of heads obtained when a fair is tossed 8 times.
Now, X is a binomial distribution with n = 8, \[p = \frac{1}{2}\]  and  \[q = 1 - \frac{1}{2} = \frac{1}{2}\] .

\[\therefore P\left( X = r \right) =^8 C_r \left( \frac{1}{2} \right)^{8 - r} \left( \frac{1}{2} \right)^r =^8 C_r \left( \frac{1}{2} \right)^8 , r = 0, 1, 2, . . . , 8\]

 

Probability of getting atleast 6 heads

 

\[= P\left( X \geq 6 \right)\]
\[ = P\left( X = 6 \right) + P\left( X = 7 \right) + P\left( X = 8 \right)\]
\[ =^8 C_6 \left( \frac{1}{2} \right)^8 +^8 C_7 \left( \frac{1}{2} \right)^8 +^8 C_8 \left( \frac{1}{2} \right)^8 \]
\[ = \left( 28 + 8 + 1 \right) \times \frac{1}{256}\]
\[ = \frac{37}{256}\]