1 1 vote An urn contains $5$ red balls and $5$ black balls. In the first draw, one ball is picked at random and discarded without noticing its colour. The probability to get a Red ball in the second draw is $\frac{1}{2} \\$ $\frac{4}{9} \\$ $\frac{5}{9} \\$ $\frac{6}{9} $ Probability & Statistics gate2017-ee-2 probability-and-statistics conditional-probability + – Arjun 1.9k views answer comment Share Follow 0 reply Please log in or register to add a comment.
3 3 votes $5 RED balls +5 BLACK balls = Total 10 balls Here we have two cases. Case -1: First ball drawn is RED and second ball drawn is also RED Case-2: First ball drawn is BLACK and second ball drawn is RED Probability of FIRST drawn ball is RED and SECOND drawn ball is RED = 5/10 * 4/9 = 20/90 Probability of FIRST drawn ball is BLACK and SECOND ball is RED =5/10* 5/9= 25/90 Required probability = 20/90 + 25/90 = 1/2 katarisoma answered Nov 12, 2018 katarisoma comment Share Follow 0 reply Please log in or register to add a comment.
