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?
Answer & explanation
Correct answer: option 1
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.