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If two sections of strengths 30 and 45 are formed from 75 students who are admitted in a school, then the probability that two particular students are always together in the same section is
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1997 Upvotes
Verified Answer
The correct answer is:
\(\frac{19}{37}\)
According to given informations, the required probability
\(\begin{aligned}
& =\frac{{ }^{73} C_{28}+{ }^{73} C_{43}}{{ }^{75} C_{30}} \\
& =\frac{\frac{73 !}{28 ! 45 !}+\frac{73 !}{43 ! 30 !}}{\frac{75 !}{30 ! 45 !}} \\
& =\frac{\frac{1}{45 \times 44}+\frac{1}{30 \times 29}}{\frac{75 \times 74}{(30 \times 29)(45 \times 44)}} \\
& =\frac{(30 \times 29)+(44 \times 45)}{(75 \times 74)} \\
\end{aligned}\)
\(\begin{aligned}
& =\frac{(2 \times 29)+(44 \times 3)}{(5 \times 74)}=\frac{29+(22 \times 3)}{5 \times 37} \\
& =\frac{29+66}{5 \times 37}=\frac{95}{5 \times 37}=\frac{19}{37}
\end{aligned}\)
Hence, option (2) is correct.
\(\begin{aligned}
& =\frac{{ }^{73} C_{28}+{ }^{73} C_{43}}{{ }^{75} C_{30}} \\
& =\frac{\frac{73 !}{28 ! 45 !}+\frac{73 !}{43 ! 30 !}}{\frac{75 !}{30 ! 45 !}} \\
& =\frac{\frac{1}{45 \times 44}+\frac{1}{30 \times 29}}{\frac{75 \times 74}{(30 \times 29)(45 \times 44)}} \\
& =\frac{(30 \times 29)+(44 \times 45)}{(75 \times 74)} \\
\end{aligned}\)
\(\begin{aligned}
& =\frac{(2 \times 29)+(44 \times 3)}{(5 \times 74)}=\frac{29+(22 \times 3)}{5 \times 37} \\
& =\frac{29+66}{5 \times 37}=\frac{95}{5 \times 37}=\frac{19}{37}
\end{aligned}\)
Hence, option (2) is correct.
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