Respuesta :
Answer:
The probability that the chosen ball came from urn B, given that it was a yellow ball is [tex]\frac{230}{533}[/tex].
Step-by-step explanation:
Given information:
Urn A : 4 yellow and 6 red balls
Urn B : 10 yellow and 5 red balls
Urn C : 11 yellow and 12 red balls
Let A,B and C are events that bag A,B and C are selected respectively.
[tex]P(A)=P(B)=P(C)=\frac{1}{3}[/tex]
E be the event that a yellow ball is drawn.
The probability that a yellow ball is drawn from urn A.
[tex]P(\frac{E}{A})=\frac{4}{10}=\frac{2}{5}[/tex]
The probability that a yellow ball is drawn from urn B.
[tex]P(\frac{E}{B})=\frac{10}{15}=\frac{2}{3}[/tex]
The probability that a yellow ball is drawn from urn C.
[tex]P(\frac{E}{C})=\frac{11}{23}[/tex]
We need to find the probability that the chosen ball came from urn B, given that it was a yellow ball.
[tex]P(\frac{B}{E})=\frac{P(B)P(\frac{E}{B})}{P(A)P(\frac{E}{A})+P(B)P(\frac{E}{B})+P(C)P(\frac{E}{C})}[/tex]
[tex]P(\frac{B}{E})=\frac{\frac{1}{3}(\frac{2}{3})}{\frac{1}{3}(\frac{2}{5})+\frac{1}{3}(\frac{2}{3})+\frac{1}{3}(\frac{11}{23})}[/tex]
[tex]P(\frac{B}{E})=\frac{230}{533}[/tex]
[tex]P(\frac{B}{E})\approx 0.43152[/tex]
Therefore the probability that the chosen ball came from urn B, given that it was a yellow ball is [tex]\frac{230}{533}[/tex].
