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JEEMAIN and NEET
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JEEMAIN PAST PAPERS
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1998
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A problem in EAMCET examination is given to 3 students A, B and C whose chance of solving it are $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$ respectively, the probability that the problem is solved, is :
( A ) $\frac{1}{4}$
( B ) $\frac{1}{24}$
( C ) $\frac{23}{24}$
( D ) $\frac{3}{4}$
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