Question

Three bags I, II and III contain 1 white, 2 black and 3 red balls; 2 white, 1 black and 1 red balls; and 4 white, 5 black and 3 red balls, respectively. One bag is selected at random and two balls are drawn. These happen to be one white and one red. Find the probability that these balls were drawn from bag I.​

Answer 1

Answer:

Bag I- 1 W, 2 B, 3 R

Bag II- 2 W, 1 B, 1 R

Bag III= 4 W, 3 B, 2 R

Probability that first bag is chosen and one is white and other red=

$\frac{1}{3}$*$\frac{1*3}{6*6}$

Probability that second bag is chosen and one is white and other red=

$\frac{1}{3} *$$\frac{4*2}{9*9}$

using Bayes rule

Probability that it came from third bag=

Half equation is in picture

Answer 2

Answer:

=> Bag l = 1 W, 2 B, 3 R.

=> Bag ll = 2 W, 1 B, 1 R.

=> Bag lll = 4 W, 3 B, 2 R.

=> Probability that first bag is chosen and one is white and other red = 1/3 × 1×3/6×6.

=> Probability that second bag is chosen and one is white and other red = 1/3 × 4×2/6×6 using Bayes rule.

=> Probability that it came from third bag = Half equation is in picture.

Step-by-step explanation:

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