From an urn containing six balls, 3 white and 3 black ones, a person selects at random an even number of balls (all the different ways of drawing an even number of balls are considered equally probable, irrespective of their number). Then the probability that there will be the same number of black and white balls among them

Question

From an urn containing six balls, 3 white and 3 black ones, a person selects at random an even number of balls (all the different ways of drawing an even number of balls are considered equally probable, irrespective of their number). Then the probability that there will be the same number of black and white balls among them (A) 4/5 (B) 11/15 (C) 11/30 (D) 2/5

Tardigrade Admin · Accepted Answer

Total number of possible cases = 3 (either 2 or 4 or 6 are drawn) Hence required probability =(1/3)(( 3 C 1 × 3 C 1/ 6 C 2)+( 3 C 2 × 3 C 2/ 6 C 4)+( 3 C 3 × 3 C 3/ 6 C 6))=(11/15) ⇒

4/5

11/15

11/30

2/5

Total number of possible cases = 3 (either 2 or 4 or 6 are drawn) Hence required probability =31​(6C2​3C1​×3C1​​+6C4​3C2​×3C2​​+6C6​3C3​×3C3​​)=1511​⇒

Total number of possible cases = 3 (either 2 or 4 or 6 are drawn)
Hence required probability $= \frac{1}{3} (\frac{^{3} C _{1} \times ^{3} C _{1}}{^{6} C _{2}} + \frac{^{3} C _{2} \times ^{3} C _{2}}{^{6} C _{4}} + \frac{^{3} C _{3} \times ^{3} C _{3}}{^{6} C _{6}}) = \frac{11}{15} \Rightarrow$