Drawing balls, combinatorics and Bayes

Problem

A box contains $6$ blue and $4$ red balls; $4$ are drawn. a) (Simultaneous draw) Probability that at least one is blue; that all are red; that $2$ are red and $2$ blue. c) (One at a time, without replacement) Probability of drawing $1$ blue followed by $3$ red. e) (One at a time, with replacement) Probability that the $1^\text{st}$ is blue and the $3^\text{rd}$ red. f) A box $B$ contains $6$ red and $4$ blue. Box $A$ is chosen with probability $0{,}86$, otherwise $B$, and one ball is drawn, which turns out blue. What is the probability it came from box $A$?

Solution

a) $P(\text{all red})=\dfrac{\binom{4}{4}}{\binom{10}{4}}=\dfrac{1}{210}\approx0{,}0048$, so $P(\ge1\text{ blue})=1-\dfrac{1}{210}=\dfrac{209}{210}\approx0{,}9952$. $\;P(2R,2B)=\dfrac{\binom{4}{2}\binom{6}{2}}{\binom{10}{4}}=\dfrac{6\cdot15}{210}=\dfrac{3}{7}\approx0{,}4286$. c) $\dfrac{6}{10}\cdot\dfrac{4}{9}\cdot\dfrac{3}{8}\cdot\dfrac{2}{7}=\dfrac{1}{35}\approx0{,}0286$. e) With replacement the draws are independent: $\dfrac{6}{10}\cdot\dfrac{4}{10}=0{,}24$. f) Bayes with $P(B\mid A)=0{,}6$, $P(B\mid \text{box }B)=0{,}4$: $P(A\mid B)=\dfrac{0{,}86\cdot0{,}6}{0{,}86\cdot0{,}6+0{,}14\cdot0{,}4}=\dfrac{0{,}516}{0{,}572}\approx0{,}9021$. $\boxed{\,a)\ 0{,}9952;\ \tfrac{1}{210};\ 0{,}4286\quad c)\ \tfrac{1}{35}\quad e)\ 0{,}24\quad f)\ 0{,}9021\,}$