MHT CET · Maths · Probability
Three urns respectively contain 2 white and 3 black, 3 white and 2 black and 1 white and 4 black balls. If one ball is drawn from each urn, then the probability that the selection contains 1 black and 2 white balls is
- A \(\frac{13}{125}\)
- B \(\frac{37}{125}\)
- C \(\frac{28}{125}\)
- D \(\frac{33}{125}\)
Answer & Solution
Correct Answer
(B) \(\frac{37}{125}\)
Step-by-step Solution
Detailed explanation
\( P(W_1) = \frac{2}{5}, P(B_1) = \frac{3}{5} \) \( P(W_2) = \frac{3}{5}, P(B_2) = \frac{2}{5} \)
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