probability solve

Question

Answer 1

This can be solved using total probability theorem.

Comments

Bayes theorem is not applicable here. For further clarification refer ocw probability Lectures by Tsitsiklis. Those are really awesome.

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I think lecture 2 about conditioning and Bayes theorem

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ok, tnks

Answer 2

I think they have got the answer like this:

Case 1)   1 white is transfrred from P to Q. Thus, Q contains  5 white and 3 black. And then 1 white is transferred from Q to P.

Case 2)   1 black is transfrred from P to Q. Thus, Q contains  4 white and 4 black. And then 1 black is transferred from Q to P.

Case 3)   1 white is transfrred from P to Q. Thus, Q contains  5 white and 3 black. And then 1 black is transferred from Q to P.

Case 4)   1 black is transfrred from P to Q. Thus, Q contains  4 white and 4 black. And then 1 white is transferred from Q to P.

Lets evaluate case 1: After the transfers are over, 3 balls white out of 7 total balls

Probablity of selecting white from P = $\frac{3}{7}$

Now, Q has 5 white balls.

Probablity of selecting white from Q = $\frac{5}{8}$

Now, probablity that after the transfers are over,

you have 3 balls white out of 7 total balls in P

= $\frac{3}{7}$ * $\frac{5}{8}$

= $\frac{15}{56}$

Lets evaluate case 2: After the transfers are over, 3 balls white out of 7 total balls

Probablity of selecting black from P = $\frac{4}{7}$

Now, Q has 4 black balls.

Probablity of selecting black from Q = $\frac{4}{8}$

Now, probablity that after the transfers are over,

you have 3 balls white out of 7 total balls in P

= $\frac{4}{7}$ * $\frac{4}{8}$

= $\frac{16}{56}$

Lets evaluate case 3: After the transfers are over, 2 balls white out of 7 total balls

Probablity of selecting white from P = $\frac{3}{7}$

Now, Q has 3 black balls.

Probablity of selecting black from Q = $\frac{3}{8}$

Now, probablity that after the transfers are over,

you have 2 balls white out of 7 total balls in P

= $\frac{3}{7}$ * $\frac{3}{8}$

= $\frac{9}{56}$

Lets evaluate case 4: After the transfers are over, 4 balls white out of 7 total balls

Probablity of selecting black from P = $\frac{4}{7}$

Now, Q has 4 white balls.

Probablity of selecting white from Q = $\frac{4}{8}$

Now, probablity that after the transfers are over,

you have 4 balls white out of 7 total balls in P

= $\frac{4}{7}$ * $\frac{4}{8}$

= $\frac{16}{56}$

Now, calculating total probablity is similar to calculating the expected value.

Now, total probablity 

= $\frac{3}{7}$ * $\frac{15}{56}$  +  $\frac{3}{7}$ * $\frac{16}{56}$ + $\frac{2}{7}$ * $\frac{9}{56}$ + $\frac{4}{7}$ * $\frac{16}{56}$

= $\frac{25}{56}$ 

Answer 3

In P probability of a ball is white=$\frac{3}{7}$

Now, a ball transferred from P to Q and then Q to P, then what can happen?

that a black ball going to Q and a white ball returns to P.

Then probability of P is $\frac{4}{7}$

 

Similar case happen for Q too.

So, at first probability of Q is $\frac{4}{7}$

and after exchanging a ball probability of Q is $\frac{5}{7}$

 

Now, put Bias Theorem

$\frac{\frac{1}{2}.(\frac{3}{7}+\frac{4}{7})}{\frac{1}{2}.(\frac{3}{7}+\frac{4}{7})+\frac{1}{2}.(\frac{4}{7}+\frac{5}{7})}$

=$\frac{7}{16}$

Ans D)

Comments

answer given is (c)

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but how not given ,rt?