Question

Six unbiased coins are tossed $6400$ times. Using Poisson distribution, the approximate probability of getting six heads $2$ times is

• A. $\frac{e^{-64}(64{)}^2}{2!}$
• B. $\frac{e^{-100}(100{)}^2}{2!}$
• C. $1-\frac{e^{-100}(100{)}^x}{x!}$
• D. $\frac{e^{-100}(100{)}^x}{x!}$

Answer 1

The correct option is B $\frac{e^{-100}(100{)}^2}{2!}$

Here $\lambda ={\left(\frac{1}{2}\right)}^6\times 64100=100$
Hence probability of getting six heads two times $=P(X=2)=\frac{e^{-100}(100{)}^2}{2!}$