Question

Suppose you have two urns with poker chips in them. Urn I contains two red chips and four white chips.Urn II contains three red chips and one white chip. You randomly select one chip from urn I and put itinto urn II. Then you randomly select a chip from urn II.What is the probability that the chip you select from urn II is red?

Answer 1

Answer:

$P(R_{2}) =\frac{10}{15} = 0.667$

Step-by-step explanation:

Step 1: Understanding the possible events

Selecting a chip from Urn I and then adding that chip to Urn II and then selecting a red chip from Urn II can be completed in two ways:

A. Selecting a red chip from Urn I and adding it to Urn II and then selecting a red chip from Urn II

B. Selecting a white chip from Urn I and adding it to Urn II and then selecting a red chip from Urn II

Therefore total probability is:

                                         $P(R_{2}) = P(A) + P(B)$

Step 2: Probability of selecting either chip from Urn I

Urn I contains 2 reds and 4 white chips, that gives a total of 6 chips.

                                             $P(R_{1}) = \frac{2}{6} =\frac{1}{3}$

                                             $P(W_{1}) = \frac{4}{6} =\frac{2}{3}$

Step 3: Probability of selecting a red chip from Urn II

Urn II originally contains 3 reds and 1 white chip, that gives a total of 4 chips.

Remember: Once a chip is added from Urn I to Urn II the total number of chips will increase in the Urn II

Case 1: When a red chip is added from Urn I to Urn II

Red chips    = 4

White chips = 1

Total Chips  = 5

                                                  $P(R_{2_1}) = \frac{4}{5}$

Case 2: When a white chip is added from Urn I to Urn II

Red chips    = 3

White chips = 2

Total Chips  = 5

                                                  $P(R_{2_2}) = \frac{3}{5}$

Therefore the total Probability of selecting a chip from Urn I and then adding that chip to Urn II and then selecting a red chip from Urn II can be calculated as:

                                          $P(R_{2}) = P(A) + P(B)$

                           $P(R_{2}) = P(R_{1}) . P(R_{2_1}) + P(W_{1}) . P(R_{2_2})$

                                           $P(R_{2}) =\frac{1}{3} . \frac{4}{5} + \frac{2}{3} .\frac{3}{5}$

                                             $P(R_{2}) =\frac{4}{15} + \frac{2}{5}$

                                            $P(R_{2}) =\frac{10}{15} = 0.667$