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Sagot :
Explanation
We are given the following:
[tex]\begin{gathered} Bag=\begin{cases}{2\text{ }blue\text{ }marbles} \\ {1\text{ }white\text{ }marbles} \\ {1\text{ }red\text{ }marbles}\end{cases} \\ Total\text{ }marbles=2+1+1=4 \end{gathered}[/tex]We are required to determine the following probabilities:
[tex]\begin{gathered} (a)\text{ }a\text{ }blue,\text{ }then\text{ }a\text{ }red \\ (b)\text{ }a\text{ }red,\text{ }then\text{ }white \\ (c)\text{ }a\text{ }blue,\text{ }then\text{ }a\text{ }blue,\text{ }then\text{ }a\text{ }blue \end{gathered}[/tex]We know that probability is calculated as:
[tex]Prob.=\frac{Number\text{ }of\text{ }required\text{ }outcome}{Number\text{ }of\text{ }possible\text{ }or\text{ }total\text{ }outcome}=\frac{n(E)}{n(S)}[/tex]For Question A:
We can determine the probability of a blue, then a red as:
[tex]\begin{gathered} P(blue\text{ }and\text{ }red)=P(Blue)\times P(Red) \\ =\frac{2}{4}\times\frac{1}{4}=\frac{2}{16}=\frac{1}{8} \\ \therefore P(blue\text{ }and\text{ }red)=\frac{1}{8} \end{gathered}[/tex]For Question B:
We can determine the probability of a red, then white as:
[tex]\begin{gathered} P(red\text{ a}nd\text{ w}h\imaginaryI te)=P(Red)\times P(Wh\imaginaryI te) \\ =\frac{1}{4}\times{}\frac{1}{4}=\frac{1}{16} \\ \operatorname{\therefore}P(red\text{ a}nd\text{ w}h\imaginaryI te)=\frac{1}{16} \end{gathered}[/tex]For Question C:
We can determine the probability of a blue, then blue, then blue as:
[tex]\begin{gathered} P(blue,blue,blue)=P(blue)\times P(blue)\times P(blue) \\ =\frac{2}{4}\times\frac{2}{4}\times\frac{2}{4}=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8} \\ \therefore P(blue,blue,blue)=\frac{1}{8} \end{gathered}[/tex]Hence, the answers are:
[tex]\begin{gathered} (a)\text{ }P(blue\text{ }and\text{ }red)=\frac{1}{8} \\ \\ (b)\text{ }P(red\text{ a}nd\text{ w}h\mathrm{i}te)=\frac{1}{16} \\ \\ (c)\text{ }P(blue,blue,blue)=\frac{1}{8} \end{gathered}[/tex]
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