Answer:
[tex]P(R_{2}) =\frac{10}{15} = 0.667[/tex]
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:
[tex]P(R_{2}) = P(A) + P(B)[/tex]
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.
[tex]P(R_{1}) = \frac{2}{6} =\frac{1}{3}[/tex]
[tex]P(W_{1}) = \frac{4}{6} =\frac{2}{3}[/tex]
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
[tex]P(R_{2_1}) = \frac{4}{5}[/tex]
Case 2: When a white chip is added from Urn I to Urn II
Red chips = 3
White chips = 2
Total Chips = 5
[tex]P(R_{2_2}) = \frac{3}{5}[/tex]
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:
[tex]P(R_{2}) = P(A) + P(B)[/tex]
[tex]P(R_{2}) = P(R_{1}) . P(R_{2_1}) + P(W_{1}) . P(R_{2_2})[/tex]
[tex]P(R_{2}) =\frac{1}{3} . \frac{4}{5} + \frac{2}{3} .\frac{3}{5}[/tex]
[tex]P(R_{2}) =\frac{4}{15} + \frac{2}{5}[/tex]
[tex]P(R_{2}) =\frac{10}{15} = 0.667[/tex]
