Question

Suppose $2$% of the people on an average are left handed. The probability of 3 or more left handed among 100 people is

• A. $3e^{-2}$
• B. $4e^{-2}$
• C. $1-5e^{-2}$
• D. $5e^{-2}$

Answer 1

Answer: (C)

$1-5e^{-2}$

Here P.D parameter $\lambda =\frac{2}{100}\times 100=2$
Hence probability that 3 or more people are left handed is $=1-P(X=0)-P(X=1)-P(X=2)$
$=1-e^{-2}-\frac{e^{-2}2}{1!}-\frac{e^{-2}{2}^2}{2!}=1-5e^{-2}$