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?
