MHTCET 22 Apr 2025 Shift 2 Paper

Concept:Probability of independent draws from three urns, adding all favorable cases.Explanation:Each urn contains 5 balls, so every single draw has denominator 5.We need exactly 1 black and 2 white balls in the selection.There are three possible positions for the black ball: from urn 1, urn 2, or urn 3.Case 1: Black from urn 1, white from urn 2, white from urn 3.$P = \frac{3}{5} \times \frac{3}{5} \times \frac{1}{5} = \frac{9}{125}$.Case 2: White from urn 1, black from urn 2, white from urn 3.$P = \frac{2}{5} \times \frac{2}{5} \times \frac{1}{5} = \frac{4}{125}$.Case 3: White from urn 1, white from urn 2, black from urn 3.$P = \frac{2}{5} \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{125}$.Required probability:$\frac{9}{125} + \frac{4}{125} + \frac{24}{125} = \frac{37}{125}$.Answer:$\frac{37}{125}$