Single Choice

In a binomial distribution $$\mathrm{B} ($$n,p $$=\displaystyle \dfrac{1}{4})$$, if the probability of at least one success is greater than or equal to $$\displaystyle \dfrac{9}{10}$$ , then $$\mathrm{n}$$ is greater than

A$$\displaystyle \dfrac{1}{\log_{10}{4}-\log_{10}{3}}$$
Correct Answer
B$$\displaystyle \dfrac{1}{\log_{10}{4}+\log_{10}{3}}$$
C$$\displaystyle \dfrac{9}{\log_{10}{4}-\log_{10}{3}}$$
D$$\displaystyle \dfrac{4}{\log_{10}{4}-\log_{10}{3}}$$

Solution

Given p= 1 4 ,q=1−p= 3 4 P(x≥1)≥ 9 10 ⇒1−P(x=0)≥ 9 10 P(x=0)=nC0p0qn=( 3 4 )n ⇒ 1 10 ≥( 3 4 )n ⇒( 3 4 )n≤ 1 10 ⇒n[log103−log104]≤−1 ⇒n≥ 1 log104−log103

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