JEE Mains · Maths · STD 12 - 13. probability
Let \(9\) distinct balls be distributed among \(4\) boxes, \(B_{1}, B_{2}, B_{3}\) and \(B_{4}\). If the probability that \(B_{3}\) contains exactly \(3\) balls is \(k\left(\frac{3}{4}\right)^{9}\) then \(\mathrm{k}\) lies in the set:
- A \(\{x \in R:|x-5| \leq 1\}\)
- B \(\{x \in R:|x-2| \leq 1\}\)
- C \(\{x \in R:|x-3|<1\}\)
- D \(\{x \in R:|x-1|<1\}\)
Answer & Solution
Correct Answer
(C) \(\{x \in R:|x-3|<1\}\)
Step-by-step Solution
Detailed explanation
\(\text { required probability }=\frac{{ }^{9} \mathrm{C}_{3} \cdot 3^{6}}{4^{9}}\) \(=\frac{{ }^{9} \mathrm{C}_{3}}{27} \cdot\left(\frac{3}{4}\right)^{9}\) \(=\frac{28}{9} \cdot\left(\frac{3}{4}\right)^{9} \Rightarrow \mathrm{k}=\frac{28}{9}\) Which satisfies \(|x-3|\,<\,1\)
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