Urn i contains five red chips and four white chips; urn ii contains four red and five white chips. two chips are drawn simultaneously from urn i and placed into urn ii. then a single chip is drawn from urn ii. what is the probability that the chip drawn from urn ii is white?

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) = = P(E₂)

P(W|E₁) =

P(W|E₂) =

So, by "Bayes theorem ", we get that

Hence, our required probability is 0.56.

answer: the answer is one

step-by-step explanation:

pemdas, in this situation there is no parenthesis or exponents so you go to multiplication, 1x1=1, then division 1/1= 1 then addition 1+1=2 then subtraction 2-1=1

the answer is:

mean: 6

mad: 2.14