Question

A box has 5 blue and 4 red balls. One ball is drawn at random and not replaced. Its colour is also not noted. Then another ball is drawn at random. What is the probability of second ball being blue?

Answer 1

Toolbox:

Let\(E_1\)=1 ball drawn is Blue B

\(E_2\)=1 ball drawn is Red R

A=2ball drawn

P(A)=P\((E_1\) )P\((A/E_1\))+P\((E_2\))P\((A/E_2\))

P\((E_1\))=P(1ball drawn is B)

=\(\large\frac{5}{9}\)

P\((E_2\))=P(1ball drawn is R)

=\(\large\frac{4}{9}\)

P\((A/E_1\))=P(2nd ball drawn is blue given 1st ball dawn is blue)

=\(\large\frac{4}{8}\)

P\((A/E_2\))=P(2nd ball drawn is blue given 1st ball drawn is red)

=\(\large\frac{5}{8}\)

P(A)=\(\large\frac{5}{9}\times\)\(\large\frac{4}{8}\)+\(\large\frac{4}{8}\times\)\(\large\frac{5}{8}\)

Answer 2

Probability =5/9*4/8+4/9*5/8=40/72=5/9