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Four persons are to be chosen at random from a group of 3 men, 2 women and 4 children. Find the probability of selecting: (i)1man  1woman and 2 children (ii)Exactly 2 children             (iii) 2 women

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Total persons=2+3+4=9
total ways=`.^9C_4`
1)`.^3C_!*.^2C_1*.^4C_2` favorable cases
`P=(.^3C_1*.^2C_1.^4C_1)/(.^9C_4)`
`P=2/7`
2)Exactly 2 children
Case 1) 2 childdren 1 men 1 women
`.^4C_2*.^3C_1*.^2C_1`
Case 2) 2 children 2 mean 0 women
`.^4C_2*.^3C_2`
Case 3) 2 children 0 mean 2 women
`.^4C_2*.^2C_2`
Total ways=`.^4C_2*.^3C_1*.^2C_1+.^4C_2.^3C_2+.^4C_2.^2C_2`
`=10/21`
Case1)`.^2C_2*.^4C_1*.^3C_1=1*4*3=12`
Case2)`.^2C_2*.^4C_2=6`
Case3)`.^2C_2*.^3C_2=1*3=3`
`P=(12+6+3)/(.^9C_4)=1/6`.

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