Two bags containing different sets of marbles.

N1CKYYY

New member
Joined
Jan 11, 2008
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3
Different people have different results when they try to solve this example. Interesting. Which one is true i have no idea. Why don't you try to solve it too so i compare your result with the others

We have 2 bags.

Bag 1 contains:

2 white marbles
3 green marbles
4 red marbles

Bag 2 contains

4 white marbles
2 green marbles
3 red marbles.

We take one marble from each bag and put them in a third one.

Questions arise:

1) What is the probability both marbles are THE SAME COLOR.

2) If we take one marble from the third bag what is the probability this marble to be WHITE COLORED.

3) If we take a marble from the third bag and it is white what is the probability the third bag to contains two white marbles.

Anyone?
 

stapel

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Staff member
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Feb 4, 2004
Messages
15,943
N1CKYYY said:
Different people have different results when they try to solve this example.
Please reply with a clear listing of the work and reasoning used to obtain each result.

Thank you! :D

Eliz.
 

N1CKYYY

New member
Joined
Jan 11, 2008
Messages
3
Work 1:

Let W1, G1 and R1 be the events that the marble selected from the first bag is white, green or red respectively.
Let W2, G2 and R2 be the events that the marble selected from the second bag is white, green or red respectively.
=> P(W1) = 2/9, P(G1) = 3/9 and P(R1) = 4/9
and P(W2) = 4/9, P(G2) = 2/9 and P(R2) = 3/9

1) Required probability
= P [ (W1?W2) U (G1?G2) U (R1?R2) ]
= P(W1?W2) + P(G1?G2) + P(R1?R2)
(because events W1?W2, G1?G2 and R1?R2 are mutually exclusive)
= P(W1)*P(W2) + P(G1)*(G2) + P(R1)*P(R2)
= (2/9)(4/9) + (3/9)(2/9) + (4/9)(3/9)
= 26/81

2) Let A, B and C represent events that one, two or none of the marbles in the third bag are white respectively.
P(A) = P(W1?W2') + P(W2?W1')
(' indicates complementary event)
=> P(A) = (2/9)*(5/9) + (4/9)*(7/9) = 38/81
P(B) = 26/81 worked in (1) as above.
P(C) = 1 - P(A) - P(B) = 17/81
If D represents the event that marble selected from bag 3 is white, then P(A/D) = (1/2) PB/D) = 1 and P(C/D) = 0
P(D) = P(A)P(D/A) + P(B)P(D/B) + P(C)P(C/A)
=> P(D) = (38/81)(1/2) + (17/81)(1) + 0 = 4/9

3) We have to find probability of the event B/D
Using Beyes' rule,
P(B/D)
= P(B)*P(D/B) / P(D)
= (17/81) / (36/81)
= 17/36

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Work 2:

1) The prob(white, white) = 2/9 * 4/9 = 8/81
The prob(green, green) = 3/9 * 2/9 = 6/81
The prob(red, red) = 4/9 * 3/9 = 12/81

Prob(same colour) = 26/81.

2) Pick a white in the third bag
prob(at least a white) = prob(white, white) + prob(white, green or red) + prob(green or red, white)
prob(at least a white) = 8/81 + 10/81 + 28/81 = 46/81

prob(white given at least a white in the bag) = (8 cases + 1/2 (46 - 8))/46 = 27/46

prob( at least a white) = (8 cases + 1/2 (46 - 8))/81 = 27/81

3) Only 8 cases are favorable out of 81: 8/81

* 1 hour ago

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Work 3:

XY = {We take from the first bag marble which is X colored and from second bag marble which is Y colored}

X,Y subset of {W, G, R} where W - white, G- green and R-red

We have

P(WW) = 8/81
P(GG) = 6/81
P(RR) = 12/81

So to solve the first question we gotta sum these P(WW) + P(GG) + P(RR) = 26/81

Now lets see the second question

lets K = {We take a white marble from the third bag}

It is easy to check that this is possible when we have WW, WG, WR, GW, RW and besides

P(K|WW) = 1
P(K|WG) = P(K|WR) = P(K|GW) = P(K|RW) = 1/2

Then we can compute P(K) = P(WW)*P(K|WW) + P(WG)*P(K|WG) + P(WR)*P(K|WR) + P(GW)*P(K|GW) + P(RW)*P(K|RW) = 32/81

Now lets see the third question

We gotta find P(WW|K) = P(WW)*P(K|WW) / P(K) = ( 8/81 ) / 32/81 = 8/32 = 1/4

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Which one is right, if any?
 
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