Urn I contains three red chips and one white chip. Urn II contains two red chips and two white chips. One chip is drawn from each urn and transferred to the other urn. Then a chip is drawn from the first urn. What is the probability that the chip ultimately drawn from urn I is red?

0.7

Step-by-step explanation:

The Probability of Chips in Urn 1

P(R) = 3/4

P(W) = 1/4

The Probability of Chips in Urn 2

P(R) = 1/2

P(W) = 1/2

The problem can be solved using probability tree method. The probability tree of the given question is attached with the answer.

The Probability tree shows the exchange of chips firstly and then the final probability of Red Chips and White Chips in Urn 1.

To find the final answer we can sum up the probabilities of the branches which are ticked in the diagram.

Probability of Branch 1 (ticked in diagram)

3/4 x 3/5 x 3/4 = 0.3375

Probability of Branch 2 (ticked in diagram)

3/4 x 2/5 x 2/4 = 0.15

Probability of Branch 3 (ticked in diagram)

1/4 x 2/5 x 1 = 0.1

Probability of Branch 4 (ticked in diagram)

1/4 x 3/5 x 3/4 = 0.1125

Final Probability = 0.3375 + 0.15 + 0.1 + 0.1125

= 0.7 Answer

correct answer:

i^31 = -i