An urn contains 5 red ball and 5 black balls. In the first draw,
| An urn contains 5 red ball 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
A. <span class="math-tex">\(\frac{1}{2}\)</span>
B. <span class="math-tex">\(\frac{4}{9}\)</span>
C. <span class="math-tex">\(\frac{5}{9}\)</span>
D. <span class="math-tex">\(\frac{6}{9}\)</span>
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Calculation:
Given:
Urn contains 5 red balls, 5 black balls.
One ball is picked at random.
Case (i): The first ball is red ball
Probability to get a red ball in the second draw is
\({P_1} = \frac{5}{{10}} \times \frac{4}{9} = \frac{2}{9}\)
Case (ii): The first ball is black ball
Probability to get a red ball in the second draw is
\({P_2} = \frac{5}{{10}} \times \frac{5}{9} = \frac{5}{{18}}\)
Required probability (P) \(= {P_1} + {P_2} = \frac{2}{9} + \frac{5}{{18}} = \frac{1}{2}\)