There are two urns with green and blue chips. Urn I contains 3 green and 2 blue

Question

There are two urns with green and blue chips. Urn I contains 3 green and 2 blue chips. Urns II 2 green and 4 blue chips. Two chips are removed from each urn. Let GI denote the number of green chip in the sample from urn I. Let GII denote the number of green chips in the sample from urn II. Determine the probability mass functions for each R.V. List the nine possible ordered pairs (GI, GII). Determine the probabilities, P ({GI, GII}) Let Z = GI + GII. Determine the probability mass function for the R.V. Z.

Explanation / Answer

Answer: Based on the information provided:

Urn I contains: 3 green and 2 blue chips

Urn II contains: 2 green and 4 blue chips

2 chips are removed from each Urn – total of 4chips are removed from both Urns

GI = Number of green chip in the sample from Urn I

GII = Number of green chips in the sample from Urn II

P(GI) = Probability(Removing green chip from Urn I)

Probability mass function for each Random Variable:

Since 2 chips have been removed from Urn I therefore the Random Variable GI (number of green chip in the sample from Urn I) can take on 3 values: 0 green, 1 green or 2 green chips

Now we have to determine the probability values for each scenario. For this we need to determine the number of ways of choosing 2 chips from total 5 chips in Urn I.

There are total 5 chips in Urn I (3 green and 2 blue chips) and we are removing 2 chips from it and there are:

nCr ways = n!/(r!*(n-r)!)

i.e. 5C2 = 5!/(2!*(5-2)!) = 10 ways

For GI = 0, the sample chip will contain 0 green chip and 2 blue chips: (3C0)*(2C2)/(5C2) = (1*1)/10 = 0.1

For GI =1, the sample chip will contain 1 green chip and 1 blue chip: (3C1)*(2C1)/5C2 = (3*2)/10 = 0.6

For GI = 2, the sample chip will contain 2 green chip and 0 blue chip: (3C2)*(2C0)/5C2 = (3*1)/10 = 0.3

These are the different values of P(GI = gi) for all possible values of GI and is the probability mass function describing the distribution of GI

Similarly for GII:

P(GI) = Probability(Removing green chip from Urn II)

Since 2 chips have been removed from Urn II therefore the Random Variable GI (number of green chip in the sample from Urn II) can take on 3 values: 0 green, 1 green or 2 green chips

Now we have to determine the probability values for each scenario. For this we need to determine the number of ways of choosing 2 chips from total 6 chips in Urn II.

There are total 6 chips in Urn II (2 green and 4 blue chips) and we are removing 2 chips from it and there are:

nCr ways = n!/(r!*(n-r)!)

i.e. 6C2 = 6!/(2!*(6-2)!) = 15 ways

For GI = 0, the sample chip will contain 0 green chip and 2 blue chips: (2C0)*(4C2)/(6C2) = (1*6)/15 = 0.4

For GI =1, the sample chip will contain 1 green chip and 1 blue chip: (2C1)*(4C1)/6C2 = (2*4)/15 = 0.533

For GI = 2, the sample chip will contain 2 green chip and 0 blue chip: (2C2)*(4C0)/6C2 = (1*1)/15 = 0.0666

These are the different values of P(GI = gi) for all possible values of GI and is the probability mass function describing the distribution of GI

Considering chips are removed without replacement

Nine Possible Ordered Pairs (GI, GII) and probabilities Urn I contains: 3 green and 2 blue chips, Urn II contains: 2 green and 4 blue chips):

(Green green, green green) = (3/5*2/4)*(2/6*1/5) = 12/600 = 0.02

(Green green, green blue) = (3/5*2/4)*(2/6*4/5) = 48/600 = 0.08

(Green green, blue blue) = (3/5*2/4)*(4/6*3/5) = 72/600 = 0.12

(Green blue, green green) = (3/5*2/4)*(2/6*1/5) = 36/600 = 0.06

(Green blue, blue green) = (3/5*2/4)*(4/6*2/5) = 48/600 = 0.08

(Green blue, blue blue) = (3/5*2/4)*(4/6*3/5) = 72/600 = 0.12

(Blue blue, blue blue) = (2/5*1/4)*(4/6*3/5) = 24/600 = 0.04

(Blue blue, green green) = (2/5*1/4)*(2/6*1/5) = 4/600 = 0.00666

(Blue green, blue green) = (2/5*3/4)*(4/6*2/5) = 48/600 = 0.08

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