Question

A large box contains a large number of identical balls of which:

25% bear the number 1 only,
15% bear the number 2 only,
5% bear both the numbers 1 and 2.

(The rest of the balls do not bear any numbers.)
One ball is drawn at random and a bystander tells us that the ball does not bear the number 1.
What is the (conditional) probability that the ball bears the number 2?

Hint: Set:

A = “Ball bears the number 1”, B = “Ball bears the number 2”.

Answer 1

Solution :

A = “Ball bears the number 1”, B = “Ball bears the number 2”.

25% bear the number 1 only, that is p(A) = 0.25 , P(Ac) = 1 - 0.25 = 0.75
15% bear the number 2 only, that is p(B) = 0.15
5% bear both the numbers 1 and 2. p(A and B) = 0.05

P(B and Ac) = P(B) - P(A and B) = 0.15 - 0.05 = 0.10

Using above information,

P(the ball bears the number 2 | the ball does not bear the number 1) = P(B | Ac )

= P(B and Ac) / P(Ac)

= 0.10 / 0.75

= 0.1333

Answer : 0.1333