Respuesta :
1) The experiment consists of picking 3 blue marbles and 3 of any other color.
Therefore, we can divide such an experiment into 6 dependent events whose probabilities are calculated below
[tex]\begin{gathered} P_1(blue)=\frac{5}{9+10+5}=\frac{5}{24} \\ P_2(blue)=\frac{4}{9+10+4}=\frac{4}{23} \\ P_3(blue)=\frac{3}{9+10+3}=\frac{3}{22} \\ P_4(notblue)=\frac{19}{9+10+2}=\frac{19}{21} \\ P_5(notblue)=\frac{18}{20} \\ P_6(notblue)=\frac{17}{19} \end{gathered}[/tex]Therefore, the probability of the whole event is
[tex]P=P_1*P_2*...*P_6=\frac{51}{14168}\approx0.0036[/tex]The answer to part 1) is 0.0036.
2) We need to pick 2 blue, 2 red, and 2 white marbles; therefore,
[tex]\begin{gathered} P_1(blue)=\frac{5}{24},P_2(blue)=\frac{4}{23} \\ P_3(red)=\frac{9}{22},P_4(red)=\frac{8}{21} \\ P_5(white)=\frac{10}{20},P_6(white)=\frac{9}{19} \\ \Rightarrow P=P_1*...*P_6=\frac{45}{33649}\approx0.00134 \end{gathered}[/tex]The answer to part 2) is 0.00134
3)
[tex]\begin{gathered} P_1(nonred)=\frac{15}{24},P_2(nonred)=\frac{14}{23},...,P_6(nonred)=\frac{10}{19} \\ \Rightarrow P=\frac{15!}{9!}*\frac{18!}{24!}=\frac{65}{1748}\approx0.03719 \end{gathered}[/tex]The answer to part 3) is 0.03719
4)
[tex]\begin{gathered} P(allblue)=0 \\ P(allred)=\frac{9!}{3!}*\frac{18!}{24!}\approx0.00062 \\ P(allwhite)=\frac{10!}{4!}*\frac{18!}{24!}\approx0.00156 \end{gathered}[/tex]The probabilities of the three subevents in part 4 are shown above.
