MHT CET · Maths · Probability
First bag contains 3 red and 5 black balls and second bag contains 6 red and 4 black balls. A ball is drawn from each bag. The probability that one ball is red and the other is black, is
- A \(\frac{41}{80}\)
- B \(\frac{21}{40}\)
- C \(\frac{3}{20}\)
- D \(\frac{3}{8}\)
Answer & Solution
Correct Answer
(B) \(\frac{21}{40}\)
Step-by-step Solution
Detailed explanation
Probability of drawing red ball from first bag \(=\frac{3}{8}\) and black ball \(=\frac{5}{8}\)
Similarly probability of drawing red ball from second bag \(=\frac{6}{10}\) and black ball \(\frac{4}{10}\)
\(\therefore\) Required probability \(=\left(\frac{3}{8} \times \frac{4}{10}\right)+\left(\frac{5}{8} \times \frac{6}{10}\right)=\frac{12+30}{80}=\frac{21}{40}\)
Similarly probability of drawing red ball from second bag \(=\frac{6}{10}\) and black ball \(\frac{4}{10}\)
\(\therefore\) Required probability \(=\left(\frac{3}{8} \times \frac{4}{10}\right)+\left(\frac{5}{8} \times \frac{6}{10}\right)=\frac{12+30}{80}=\frac{21}{40}\)
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