Answer :
Answer:
(a) 0.0520
(b) 0.3699
Step-by-step explanation:
Let [tex]E_{1},E_{2},E_{3}[/tex] be the events of selecting urn 1, urn 2 and urn 3 respectively.
So, [tex]P(E_{1})=P(E_{2})=P(E_{3})=\frac{1}{3}[/tex]
Let W be the event of drawing a white ball.
Now, probability of drawing 2 white balls from urn 1 is given as:
[tex]P(W\cap W \cap E_{1})=P(E_{1})\times P((W\cap W)/E_{1})\\P(W\cap W \cap E_{1})=\frac{1}{3}\times (\frac{3}{10})^{2}=\frac{9}{300}[/tex]
Probability of drawing 2 white balls from urn 2 is given as:
[tex]P(W\cap W \cap E_{2})=P(E_{2})\times P((W\cap W)/E_{2})\\P(W\cap W \cap E_{2})=\frac{1}{3}\times (\frac{8}{10})^{2}=\frac{64}{300}[/tex]
Probability of drawing 2 white balls from urn 3 is given as:
[tex]P(W\cap W \cap E_{3})=P(E_{2})\times P((W\cap W)/E_{2})=\frac{1}{3}\times 1=\frac{1}{3}[/tex]............. (as all balls are white only)
Now, probability of drawing 2 white balls is the sum of all the above probabilities and is given as:
[tex]P(2W)=P(W\cap W \cap E_{1})+P(W\cap W \cap E_{2})+P(W\cap W \cap E_{3})\\P(2W)=\frac{9}{300}+\frac{64}{300}+\frac{1}{3}\\P(2W)=\frac{9}{300}+\frac{64}{300}+\frac{100}{300}=\frac{9+64+100}{300}=\frac{173}{300}[/tex]
(a)
Probability of selecting urn 1 given that 2 white balls are drawn is:
[tex]P(E_{1}/2W)=\frac{P(E_{1})\times P(2W/E_{1})}{P(2W)}\\P(E_{1}/2W)=\frac{\frac{9}{300}}{\frac{173}{300}}=\frac{9}{173}=0.0520[/tex]
Therefore, probability of selecting urn 1 given that 2 white balls are drawn is 0.0250.
(b)
Probability of selecting urn 2 given that 2 white balls are drawn is:
[tex]P(E_{2}/2W)=\frac{P(E_{2})\times P(2W/E_{2})}{P(2W)}\\P(E_{2}/2W)=\frac{\frac{64}{300}}{\frac{173}{300}}=\frac{64}{173}=0.3699[/tex]
Therefore, probability of selecting urn 2 given that 2 white balls are drawn is 0.3699