Rom

\(\dbinom{10}{5}5! = 30240\)

\(\text{This is the same as the probability you've eaten 2 chocolates in 6 candies, and eat the 3rd chocolate on the 7th}\\ P[\text{eaten 2 chocolates in 6 candies}] = \dfrac{\dbinom{3}{2}\dbinom{5}{4}}{\dbinom{8}{6}} = \dfrac{15}{28}\\ P[\text{7th candy eaten is chocolate |2 chocolates already eaten}] = \dfrac 1 2\\ P = \dfrac{15}{28}\dfrac 1 2 = \dfrac{15}{56}\)

\(\dbinom{9}{5} = \dbinom{9}{4} = 126\)

\(P[\text{tuberculosis|positive test}] = \dfrac{P[\text{positive test|tuberculosis}]P[\text{tuberculosis}]}{P[\text{positive test}]} = \\ \dfrac{0.9\cdot 0.08}{0.9 \cdot 0.08 + 0.05\cdot 0.92} \approx 0.939948 \\ \)

\(a \dfrac{1-r^6}{1-r} = 91 = 13 \cdot 7 = 13 a\dfrac{1-r^2}{1-r}\\ \dfrac{1-r^6}{1-r^2}=13\\ r^6-13r^2+12=0\\ r=\pm 1,~\pm \sqrt{3}\\ \text{We only need consider $r=\sqrt{3}$}\\ a \dfrac{1-3}{1-\sqrt{3}}=7\\ a = \dfrac 7 2 (\sqrt{3}-1)\\ a \dfrac{1-r^4}{1-r} = 28\)

\(p=0.001\\ a)~P[\text{no one in 10000 has blood type}]=\left(1-0.001\right)^{10000} = \\ (0.999)^{10000} = 0.0000451733\\~\\ b)~P[\text{at least 1 w/type in $n$ tested}] = \\ 1-P[\text{none w/type in $n$ tested}]=\\ 1-(1-0.001)^n = 1-(0.999)^n \geq \dfrac 1 2\\ \dfrac 1 2 \geq (0.999)^n\\ \ln(1/2)\geq n \ln(0.999)\\ \dfrac{\ln(0.5)}{\ln(0.999)} \leq n\\ n \geq 693\)

\(P[\text{draw red from B}] = \\ P[\text{draw red from B|drew red from A}]P[\text{drew red from A}]+\\ P[\text{draw red from B|drew white from A}]P[\text{drew white from A}] = \\ \dfrac 5 8\dfrac 3 5+\dfrac 4 8 \dfrac 2 5 = \dfrac{23}{40}\)

we're going to need the figure.

what direction does the river flow in?

\(P[\text{emergency|alarm}] = \\ \dfrac{P[\text{alarm|emergency}]P[\text{emergency}]}{P[\text{alarm}]} = \\ \dfrac{0.95 \cdot 0.004}{0.95\cdot 0.004 + 0.02 \cdot 0.996} = 0.160202\)

\(\text{The probability the stronger team wins is the series, }\\ \text{is the sum of the probabilities that they win 3 games out of 3,4,5,6}\\ \text{times the probability they get their fourth win}\\ p = \left(\sum \limits_{n=3}^6 \dbinom{n}{3}(0.6)^3 (0.4)^{n-3}\right)(0.6) = \dfrac{11097}{15625} \approx 0.71\)