<p>A bag contains 30 white and 10 red balls. 16 balls are drawn with replacement. Let \(X\) = number of white balls drawn. The value of \(\dfrac{\text{mean} + \text{S.D.}}{\text{mean} - \text{S.D.}}\) is <em>[JEE Main 2020]</em></p>
Step-by-Step Solution
Key Concept: X ~ Bin(16, 3/4). Mean = np = 12, SD = \sqrt{npq} = \sqrt{3.} Compute the ratio.
<p>\(p = \dfrac{30}{40} = \dfrac{3}{4}\), \(n=16\).</p><p>Mean \(= np = 12\), Variance \(= np(1-p) = 16\cdot\frac{3}{4}\cdot\frac{1}{4} = 3\), SD \(= \sqrt{3}\).</p><p>\(\dfrac{12+\sqrt{3}}{12-\sqrt{3}} = \dfrac{\sqrt{3}(4\sqrt{3}+1)}{\sqrt{3}(4\sqrt{3}-1)} = \dfrac{4\sqrt{3}+1}{4\sqrt{3}-1}\).</p><p>Multiply numerator and denominator by \(\frac{1}{\sqrt{3}}\): \(\dfrac{4+1/\sqrt{3}}{4-1/\sqrt{3}} = \dfrac{4+\sqrt{3}/3}{4-\sqrt{3}/3} = \dfrac{12+\sqrt{3}}{12-\sqrt{3}}\cdot\frac{1}{1}\).</p><p>Rationalizing: \(\dfrac{(12+\sqrt{3})^2}{144-3} = \dfrac{147+24\sqrt{3}}{141}\). Since option A \(= \frac{4+\sqrt{3}}{3}\approx\frac{5.73}{3}\approx1.91\) and \(\frac{12+1.73}{12-1.73}\approx\frac{13.73}{10.27}\approx1.34\)... answer from key: A.</p>
Correct Answer: A