Answer: Our required probability is 0.56.
Step-by-step explanation:
Since we have given that
In urn I :
Number of red chips = 5
Number of white chips = 4
In urn 2 :
Number of red chips = 4
Number of white chips = 5
Two chips are drawn simultaneously from urn I and placed into urn II. Then a single chip is drawn from urn II.
Probability that the chip drawn from urn II is white is given by
P(E₁) = [tex]\dfrac{1}{2}[/tex] = P(E₂)
P(W|E₁) = [tex]\dfrac{4}{9}[/tex]
P(W|E₂) = [tex]\dfrac{5}{9}[/tex]
So, by "Bayes theorem ", we get that
[tex]P(E_2|W)=\dfrac{0.5\times \dfrac{5}{9}}{0.5\times \dfrac{4}{9}+0.5\times \dfrac{5}{9}}\\\\P(E_2|W)=\dfrac{5}{9}=0.56[/tex]
Hence, our required probability is 0.56.
