Question

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.

Answer 1

Answer:

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