A) Probability that the chosen ball is green, given that the selected bag is \[{{\text{B}}_{\text{3}}}\] equals \[\frac{3}{8}\]
B) Probability that the selected bag is \[{{\text{B}}_{\text{3}}}\] and the chosen ball is green equals \[\frac{3}{10}\]
C) Probability that the selected bag is \[{{\text{B}}_{\text{3}}}\] and the chosen ball is green equals \[\frac{5}{13}\]
D) Probability that the chosen ball is green equals \[\frac{39}{80}\]
Correct Answer: A , D
Solution :
\[\,\,\,\,\] |
\[\text{P}({{\text{B}}_{1}})\,=\,\frac{3}{10}\left| \text{P}({{\text{B}}_{2}})=\frac{3}{10} \right|\text{P}({{\text{B}}_{3}})=\frac{4}{10}\] |
[a] \[\text{P}({{\text{G}}_{1}}\left| {{\text{B}}_{3}} \right.)\,=\,\frac{3}{8}=\] |
[b] \[\text{P}({{\text{B}}_{3}}\left| \text{G} \right.)=\] |
[c] \[\text{P}({{\text{B}}_{3}}\left| \text{G} \right.)=\frac{12}{39}=\] |
[d] \[\text{P}(\text{G})=\frac{3}{10}.\frac{5}{10}+\frac{3}{10}.\frac{5}{8}+\frac{4}{10}.\frac{3}{8}\] |
\[=\frac{12+15+12}{80}=\] |
Solution :
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