Greedy Pig

At the begin. the prob. of getting a choc. candy from jar is 4/10. and of strawberry candy is 6/10.now pig ate 3 candies.it means theoritically it ate 1.2 choc. candy and 1.8 strawberry candy so 4-1.2=2.8 and 6-1.8=4.2 so now total 2.8+4.2 =7 candies r there of which 2.8 are choc. candies so prob of choc.candy is 2.8/7=0.4.

This is a conditional probability problem. First we find out the probabilities q(n) that the greedy pig has taken n (0, 1, 2 or 3) chocolate candies. They are given by:

$q(0)=\frac { { C }_{ 3 }^{ 6 } }{ { C }_{ 3 }^{ 10 } } =\frac { 1 }{ 6 }$ $q(1)=\frac { { { C }_{ 1 }^{ 4 }C }_{ 2 }^{ 6 } }{ { C }_{ 3 }^{ 10 } } =\frac { 1 }{ 2 }$ $q(2)=\frac { { { C }_{ 2 }^{ 4 }C }_{ 1 }^{ 6 } }{ { C }_{ 3 }^{ 10 } } =\frac { 3}{ 10 }$ $q(3)=\frac { { C }_{ 3 }^{ 4 } }{ { C }_{ 3 }^{ 10 } } =\frac { 1 }{ 30}$

Please note that:

$q(0)+q(1)+q(2)+q(3)=1$

Now we find out the probabilities p(n) the candy you pick is chocolate for each case of n or for (4-n) chocolate candies left. They are given by:

$p(n)=\frac { { C }_{ 1 }^{ 4-n } }{ { C }_{ 1 }^{ 7 } } =\frac { 4-n }{ 7 }$

The probability p that the candy you pick is chocolate is:

$p=p(0)q(0)+p(1)q(1)+p(2)q(2)+p(3)q(3)$ $=\frac { 4 }{ 7 } \times \frac { 1 }{ 6 } +\frac { 3 }{ 7 } \times \frac { 1 }{ 2 } +\frac { 2 }{ 7 } \times \frac { 3 }{ 10 } +\frac { 1 }{ 7 } \times \frac { 1 }{ 30 }$ $=\frac {2}{5}=\boxed{0.4}$

The stochastic systems are memory less systems, so we are not interested in the event that happened before the experiment. Even the Pig has eaten 3 Candies, the prob of getting Choc. Candy is 4/10 = .4.

Sorry there was mistake..... I corrected it.