Probability 30421

There are 25 students in the class, 12 of whom are not ready for math. There are five students in the math class. What is the probability of at least 3 being math-ready?

Correct answer:

p =  0.4625

Step-by-step explanation:

$$\begin{gathered} p_1=\frac{12}{25}=0.48 \\ {q}_1=1-p_1=1-\frac{12}{25}=\frac{1\cdot 25}{25}-\frac{12}{25}=\frac{25}{25}-\frac{12}{25}=\frac{25-12}{25}=\frac{13}{25}=0.52 \\ {C}_3(5)=\left(\genfrac{}{}{0ex}{}{5}{3}\right)=\frac{5!}{3!(5-3)!}=\frac{5\cdot 4}{2\cdot 1}=10 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot p_1^3\cdot q_1^2=10\cdot {\frac{12}{25}}^3\cdot {\frac{13}{25}}^2\doteq 0.299 \\ {C}_4(5)=\left(\genfrac{}{}{0ex}{}{5}{4}\right)=\frac{5!}{4!(5-4)!}=\frac{5}{1}=5 \\ {p}_4=\left(\genfrac{}{}{0ex}{}{5}{4}\right)\cdot p_1^4\cdot q_1^1=5\cdot {\frac{12}{25}}^4\cdot {\frac{13}{25}}^1\doteq 0.138 \\ {C}_5(5)=\left(\genfrac{}{}{0ex}{}{5}{5}\right)=\frac{5!}{5!(5-5)!}=\frac{1}{1}=1 \\ {p}_5=\left(\genfrac{}{}{0ex}{}{5}{5}\right)\cdot p_1^5\cdot q_1^0=1\cdot {\frac{12}{25}}^5\cdot {\frac{13}{25}}^0\doteq 0.0255 \\ p=p_3+p_4+p_5=0.299+0.138+0.0255=0.4625 \end{gathered}$$

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