In a binomial distribution \[B\left( {n,p = {1 \over 4}} \right),\] if the probability of at least one success is greater than or equal to \[{9 \over {10}},\] then n is greater than :
Solutions · 1
Official solution
Given that, no of trials = n
Probability of success (p) = \[{{1 \over 4}}\]
\[\therefore\] Probability of no success = 1 - \[{{1 \over 4}}\] = \[{{3 \over 4}}\]
As we know, probability of at least one success = 1 - probability of no success
\[\therefore\] According to the question,
1 - (Probability of no success in n trials) \[\ge\] \[{9 \over {10}}\]
\[ \Rightarrow \] 1 - P(x = 0) \[\ge\] \[{9 \over {10}}\]
\[ \Rightarrow \] 1 - nC0(41)0\[{\left( {{3 \over 4}} \right)\^n}\] \[\ge\] \[{9 \over {10}}\]
\[ \Rightarrow \] 1 - \[{\left( {{3 \over 4}} \right)^n}\] \[\ge\] \[{9 \over {10}}\]
\[ \Rightarrow \] \[{\left( {{3 \over 4}} \right)^n}\] \[\le\] \[{1 \over {10}}\]
\[ \Rightarrow \] \[{\left( {{4 \over 3}} \right)^n}\] \[\ge\] 10
[Taking log on both sides]
\[ \Rightarrow \] \[n\left( {\log _{10}^4 - \log _{10}^3} \right)\] \[\ge\] \[{\log _{10}^{10}}\]
\[ \Rightarrow \] \[n\left( {\log _{10}^4 - \log _{10}^3} \right)\] \[\ge\] 1
\[ \Rightarrow \] n \[\ge\] \[{1 \over {\left( {\log _{10}^4 - \log _{10}^3} \right)}}\]
\[\therefore\] Option (D) is correct.