The contents of urns i, ii, iii are as follows: 1) 1 white, 2 black and 3 red balls 2) 2 white, 1 black and 1 red balls 3) 4 white, 5 black and 3 red balls one urn is chosen at random and two balls drawn. They happen to be a white and a red. What is the probability that they come from: a. Urn i b. Urn ii
Answers
Suppose that e1, e2 and e3 would be the events of selecting urns 1, 2 and 3 respectively.
White Black Red
1) 1 2 3
2) 2 1 1
3) 4 5 3
Suppose "a" would be the event that drawn balls are, 1 red and 1 white.
P(e1) = 1/3
P(e2) = 1/3
P(e3) = 1/3
The solution of above is given in the following image.
Answer:let E1 be the event that urn I is chosen.
Let E2 be the event that urn ll is chosen.
Let E3 be 5ge event that urn lll is chosen.
Let A be the event that the two balls taken from selected urn is red and white.
P(E1)=P(E2)=P(E3)=1/3
P(A/E1)=(3×1)/6C2
=3/(6!/4!2!)
=3/{(6×5×4!)/4!2}
=6/30
=1/5
P(A/E2)=(1×2)/4C2
=4/12
=1/3
P(A/E3)=(3×4)/12C2
=2/11
P(E1/A)=33/118
P(E2/A)=55/118
P(E3/A)=30/118.
Explanation: