probability

If P(A)=probability of getting the bag chosen for drawing balls when all the 5 balls in the bag are white.

P(B)=probability of getting the bag chosen for drawing balls when only 4 balls in the bag are white

P(C)=probability of getting the bag chosen for drawing balls when only 3 balls in the bag are white

P(D)=probability of getting the bag chosen for drawing balls when only 2 balls in the bag are white

P(E)=probability of getting two white balls drawn from the bag

P(E/A)=probability of two white balls drawn when the bag is chosen which has all white balls in it

P(E/B)=probability of two white balls drawn when the bag is chosen which has only 4 white balls in it

P(E/C)=probability of two white balls drawn when the bag is chosen which has only 3 white balls in it

P(E/D)=probability of two white balls drawn when the bag is chosen which has only 2 white balls in it.

So using baye's theorem

required probability $= \frac{P(A).P(E/A)} {P(A).P(E/A)+P(B).P(E/B)+P(C).P(E/C)+P(D).P(E/D)}$

$= \frac{\frac{1}{4}.1}{\frac{1}{4}.1+\frac{1}{4}.\frac{6}{10}+ \frac{1}{4}\frac{3}{10}+\frac{1}{4}\frac{1}{10}}$ $=\boxed{\frac{1}{2}}$

Confusing even with the crutch of computer code!

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from fractions import Fraction

bags=[{'B':k,'W':5-k} for k in range(6)]

total=Fraction(0)
probs=[]
for bag in bags:
    prob=1
    for selection in range(2):
        if bag['W']<=0:
            prob=Fraction(0)
            break
        prob*=Fraction(bag['W'],sum(bag.values()))
        bag['W']+=-1
    prob=Fraction(1,5)*prob
    total+=prob
    probs.append(prob)

print probs[0]/sum(probs)

IM NOT SURE IF THIS IS EVEN A VALID SOLUTION. I think I got it just by pure luck. Ok, we start by taking probabilities of choosing two white balls if the bag contains :5 white balls, 4 white balls, 3 white balls, or two white balls:

$10/10,6/10,3/10,1/10$ respectively

Now, taking the average probability, we find it to be $1/2$

Therefore, (1/2 change picking two whites)*(10/10 chance of picking a white ball if all are white)=5/10 chance of having all whites.

Again, I'm pretty sure that Sandeep's solution is correct but I'm not sure about mine.