Questions
Five different games are to be distributed among 4 children randomly. The probability that each child get at least one game is
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a
14
b
1564
c
2164
d
None of these
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Correct option is B
Total number of ways of distribution is 45∴ n(S)=45Total number of ways of distribution so that each child gets at least one game is 45−4C135+4C225−4C3=1024−4×243+6×32−4= n(E)=240Therefore, the required probability is n(E)n(S)=24045=1564Similar Questions
The odd against an event are 5 to 2 and the odds in favour of another disjoint event are 3 to 5. Then the probability that atleast one of the events will happen is
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