MATH SOLVE

2 months ago

Q:
# There are three urns that each contain 10 balls. The first contains 3 white and 7 red balls, the second 8 white and 2 red, and the third all 10 white. Each urn has an equal probability of being selected. An urn is selected and a ball is drawn and replaced, and then another ball is drawn from the same urn. Suppose both are white. What are the following probabilities? (Round your answers to four decimal places.)(a) The first urn was selected.(b) The second urn was selected.

Accepted Solution

A:

Answer:(a) 0.0520(b) 0.3699Step-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