AP EAMCET · Maths · Probability
There are 8 boys and 7 girls in a class room. If the names of all those children are written on paper slips and 3 slips are drawn at random from them, then the probability of getting the names of one boy and two girls or one girl and two boys is
- A \(\frac{1}{5}\)
- B \(\frac{3}{4}\)
- C \(\frac{4}{5}\)
- D \(\frac{1}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{4}{5}\)
Step-by-step Solution
Detailed explanation
Total children = \(8 + 7 = 15\) Total ways to choose 3 children = \(\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455\) Ways for (1 boy and 2 girls) = \(\binom{8}{1} \times \binom{7}{2} = 8 \times \frac{7 \times 6}{2 \times 1} = 8 \times 21 = 168\) Ways…
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