A bag contains 49 balls of three different colors viz. red, orange and pink. The ratio of red balls to orange balls is 3 : 4, respectively and probability of choosing a pink ball is \(\frac{3}{7}\). If two balls are picked from the bag, then what is the probability that one ball is orange and one ball is pink?

\(\frac{21}{98}\)

Let, the number of pink balls be P

Probability of choosing a pink ball = \(\frac{P}{49}\)

⇒ \(\frac{3}{7}\) = \(\frac{P}{49}\)

P = 21

So, remaining number of balls = (49 - 21) = 28

Number of orange balls = \(\frac{3}{3\;+\;4}\) × 28= 12

Therefore, reqd. probability =  \(\frac{^{12} \mathrm{ C }_1 \times ^{21} \mathrm{ C }_1}{^{49} \mathrm{ C }_2}\)

= \(\frac{12\;×\;21}{49\;×\;\frac{48}{2}}\) = \(\frac{21}{98}\)

Hence, option A is correct.