TS EAMCET · Maths · Probability
The mean and standard deviation of a binomial variate \(X\) are 4 and \(\sqrt{3}\) respectively. Then \(P(X \geq 1)\) is equal to
- A \(1-\left(\frac{1}{4}\right)^{16}\)
- B \(1-\left(\frac{3}{4}\right)^{16}\)
- C \(1-\left(\frac{2}{3}\right)^{16}\)
- D \(1-\left(\frac{1}{3}\right)^{16}\)
Answer & Solution
Correct Answer
(B) \(1-\left(\frac{3}{4}\right)^{16}\)
Step-by-step Solution
Detailed explanation
Mean \(=n p=4\), variance \(=n q p=3\) On solving, we get \(q=\frac{3}{4}, n=16, p=\frac{1}{4}\) Now, \(\begin{aligned} p(X \geq 1) & =1-p(X=0)=1-{ }^n C_0 p^0 q^{n-0} \\ & =1-\left(\frac{3}{4}\right)^{16}\end{aligned}\)
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