Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is m/n, where gcd(m, n)=1, then m+n is equal to :

Solution

Maths sir · 3 steps

Split 'second ball black' by what came out first, then keep only the part where the first was black.

1. Set up Bayes. Let B₁ be 'first ball black' and B₂ be 'second ball black'. We need P(B₁ | B₂) = P(B₁ ∩ B₂)/P(B₂).

   First ball	Probability of this first ball	Then second is black	Joint probability
   Black	6/10	5/9	30/90
   White	4/10	6/9	24/90

2. Total for second black. Add both routes: P(B₂) = 30/90 + 24/90 = 54/90. This equals 6/10, as it should, because without information about the first draw the second ball is just like the first.

   P(B₁ | B₂) = (30/90)/(54/90) = 30/54 = 5/9

3. So. m = 5 and n = 9, which share no common factor, so m + n = 14, option (b).

The idea to keep. For 'first given second' questions, list every way the second event can happen, then divide the favourable route by the total. Without replacement, the unconditional chance for the second draw equals the first.

Answer from the NTA final answer key. Solution by PrepWiser.