MHT CET · Maths · Probability
A bag contains 5 red balls and 3 green balls. A ball is selected at random and not replaced. A second ball is then selected. The probability of selecting one red ball and one green ball is
- A \(\frac{15}{28}\)
- B \(\frac{15}{64}\)
- C \(\frac{15}{56}\)
- D \(\frac{15}{112}\)
Answer & Solution
Correct Answer
(A) \(\frac{15}{28}\)
Step-by-step Solution
Detailed explanation
\(\text {Required probability}=P(R) \cdot P\left(\frac{G}{R}\right)+P(G) \cdot\) \(P\left(\frac{R}{G}\right)\)
\(=\frac{5}{8} \cdot \frac{3}{7}+\frac{3}{8} \cdot \frac{5}{7}\)
\(=\frac{30}{56}=\frac{15}{28}\)
\(=\frac{5}{8} \cdot \frac{3}{7}+\frac{3}{8} \cdot \frac{5}{7}\)
\(=\frac{30}{56}=\frac{15}{28}\)
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