Question

# Bag 1 contains six red balls, seven blue balls, and three green balls. Bag 2 contains...

Bag 1 contains six red balls, seven blue balls, and three
green balls. Bag 2 contains eight red balls, eight blue
balls, and two green balls. Bag 3 contains two red balls,
nine blue balls, and eight green balls. Bag 4 contains
four red balls, seven blue balls, and no green balls.
Bag 1 is chosen with a probability of 0.15, bag 2 with a
probability of 0.20, bag 3 with a probability of 0.35, and
bag 4 with a probability of 0.30, and then a ball is chosen
at random from the bag. Calculate the probabilities:
(a) A blue ball is chosen.
(b) Bag 4 was chosen if the ball is green.
(c) Bag 1 was chosen if the ball is blue.

a)P(blue ball)=P(bag 1 and blue ball)+P(bag 2 and blue ball)+P(bag 3 and blue ball)+P(bag 4 and blue ball)

=0.15*(7/16)+0.2*(8/18)+0.35*(9/19)+0.3*(7/11)=0.5112

b)

P(green ball)=P(bag 1 and green ball)+P(bag 2 and green ball)+P(bag 3 and green ball)+P(bag 4 and green ball)

=0.15*(3/16)+0.2*(2/18)+0.35*(8/19)+0.3*(0/11)=0.1977

hence P(Bag 4 was chosen if the ball is green)=P(bag 4 and green ball)/P(green ball)=0.3*(0/11)/0.1977 =0

c)

P( Bag 1 was chosen if the ball is blue)=P(bag 4 and blue ball)/P(blue ball)=0.15*(7/16)/0.5112=0.1284

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