Question

In: Statistics and Probability

Suppose 4 blue and 4 red chips are in a hat. Each time we draw a...

Suppose 4 blue and 4 red chips are in a hat. Each time we draw a chip we look at its color. If it is blue, we replace it along with one new blue chip. If it is red, we replace it along with two new red chips. What is the probability that, in successive drawing of chips, the second one is blue?

Expert Solution

Number of blue chips in the hat = 4

Number of red chips in the hat = 4

Total number of chips in the hat = 4+4=8

B1 : Event of drawing a blue chip in the first draw

P(B1) = Number of blue chips in the hat / Total number of chips in the hat = 4/8 =1/2

R1 : Event of drawing a red chip in the first draw

P(R1) = Number of red chips in the hat / Total number of chips in the hat = 4/8 =1/2

B2 : Event of drawing a blue in the second draw.

Probability that, in successive drawing chips, the second one is blue = P(B2)

B2 : Event of drawing a blue in the second draw can happen :

Draw a blue chip in the first draw and draw blue chip in the second draw or Draw a red chip in the first draw and draw blue chip in the second draw

i.e (B1 and B2) or (R1 and B2)

Therefore,

P(B2) = P(B1 and B2) + P(R1 and B2) = P(B1)P(B2|B1) + P(R1)P(B2|R1)

P(B2|B1) = Probability of drawing a blue chip in the second draw given that a blue chip was drawn in the first draw:

When a blue chip is drawn in the first, it was replaced along with one new blue chip, Therefore number blue chips = 4+1 =5; Number of chips would become = 5 blue + 4 red =9

P(B2|B1) = Number blue chips now in the hat / total number of chips in the hat = 5/9

Similarly,

P(B2|R1) = Probability of drawing a blue chip in the second draw given that a red chip was drawn in the first draw:

When a red chip is drawn in the first, it was replaced along with two new red chips, Therefore number blue chips = 4; Number of red chips = 4+2 = 6 Number of chips would become = 4 blue + 6 red =10

P(B2|R1) = Number blue chips now in the hat / total number of chips in the hat = 4/10=2/5

P(B1)P(B2|B1) = (1/2) x (5/9) = 5/18

P(R1)P(B2|R1) = (1/2) x (4/10) = 4/20 = 1/5

P(B2) = P(B1)P(B2|B1) + P(R1)P(B2|R1) = 5/18 + 1/5 = 43/90
Probability that, in successive drawing chips, the second one is blue = 43/90

orchestra answered 2 years ago

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