Left handed

It is known that 25% of the population is left-handed. What is the probability that there is a maximum of three left-handers at a seminar with 30 participants?

Correct answer:

p =  3.7449 %

Step-by-step explanation:

$$\begin{gathered} q=25/100=\frac{1}{4}=0.25 \\ r=1-q=1-0.25=\frac{3}{4}=0.75 \\ {C}_0(30)=\left(\genfrac{}{}{0ex}{}{30}{0}\right)=\frac{30!}{0!(30-0)!}=\frac{1}{1}=1 \\ {p}_0=\left(\genfrac{}{}{0ex}{}{30}{0}\right)\cdot q^0\cdot r^{30-0}=1\cdot 0.25^0\cdot 0.75^{30-0}\doteq 0.0002 \\ {C}_1(30)=\left(\genfrac{}{}{0ex}{}{30}{1}\right)=\frac{30!}{1!(30-1)!}=\frac{30}{1}=30 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{30}{1}\right)\cdot q^1\cdot r^{30-1}=30\cdot 0.25^1\cdot 0.75^{30-1}\doteq 0.0018 \\ {C}_2(30)=\left(\genfrac{}{}{0ex}{}{30}{2}\right)=\frac{30!}{2!(30-2)!}=\frac{30\cdot 29}{2\cdot 1}=435 \\ {p}_2=\left(\genfrac{}{}{0ex}{}{30}{2}\right)\cdot q^2\cdot r^{30-2}=435\cdot 0.25^2\cdot 0.75^{30-2}\doteq 0.0086 \\ {C}_3(30)=\left(\genfrac{}{}{0ex}{}{30}{3}\right)=\frac{30!}{3!(30-3)!}=\frac{30\cdot 29\cdot 28}{3\cdot 2\cdot 1}=4060 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{30}{3}\right)\cdot q^3\cdot r^{30-3}=4060\cdot 0.25^3\cdot 0.75^{30-3}\doteq 0.0269 \\ p=100\cdot (p_0+p_1+p_2+p_3)=100\cdot (0.0002+0.0018+0.0086+0.0269)=3.7449\% \end{gathered}$$

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