Suppose you have 3 jars with the following contents. Jar 1 has 1 white ball and 4 black balls. Jar 2 has 2 white balls and 1 black ball. Jar 3 has 2 white balls and 1 black ball. One jar is to be? selected, and then 1 ball is to be drawn from the selected jar. The probabilities of selecting the? first, second, and third jars are? 1/2?,?1/3?, and?1/6 respectively. Find the probability the ball was drawn from Jar 2?, given that the ball is white.

The probability of finding the white ball from second Jar = 20/39

Step-by-step explanation:

Jar 1 contains 1 White ball and 4 black balls

P (w | J₁) = 1/5

Jar 2 contains 2 White balls and 1 black ball

P (w | J₂) = 2/3

Jar 1 contains 2 White balls and 1 black ball

P (w | J₃) = 2/3

Also, Given

P (J₁) = 1/2

P (J₂) = 1/3

P (J₃) = 1/6

P (w) = P (J₁) * P (w | J₁) + P (J₂) * P (w | J₂) + P (J₃) * P (w | J₃)

P (w) = 1/2*1/5 + 1/3*2/3 + 1/6*2/3 = 13/30

To find: P (J₂ | w)

According to conditional probability,

P (J₂ | w) = P (J₂) * P (w | J₂) / P (w)

P (J₂ | w) = (1/3*2/3) / (13/30)

P (J₂ | w) = (1/3*2/3) / (13/30) = 20/39

The probability of finding the white ball from second Jar = 20/39