Question

In a binomial distribution$B\left( {n,p = \dfrac{1}{4}} \right)$, if the probability of at least one success is greater than or equal to $\dfrac{9}{{10}}$ then is greater than-
$
  {\text{A}}{\text{.}}\dfrac{1}{{{{\log }_{10}}4 - {{\log }_{10}}3}} \\
  {\text{B}}{\text{.}}\dfrac{1}{{{{\log }_{10}}4 + {{\log }_{10}}3}} \\
  {\text{C}}{\text{.}}\dfrac{9}{{{{\log }_{10}}4 - {{\log }_{10}}3}} \\
  {\text{D}}{\text{.}}\dfrac{4}{{{{\log }_{10}}4 - {{\log }_{10}}3}} \\
$

Answer 1

Hint: As we are given with P(at least one) use [1 – P(none)]. Apply binomial distribution formula for probability of favorable success, and form the required inequality.

Complete step-by-step answer:
We are given with-
p = $\dfrac{1}{4}$ if p is given then q = [1-p] = $1 - \dfrac{1}{4} = \dfrac{3}{4}$

It is given that $P\left( {{\text{at least one}}} \right) \geqslant \dfrac{9}{{10}}$ which is equivalent to-
$1 - P\left( {none} \right) \geqslant \dfrac{9}{{10}}$
The binomial distribution for 0 success would be given by-
${}^n{C_0}{p^0}{q^{n - 0}} = 1.1.{\left( {\dfrac{3}{4}} \right)^n}$
As per the given condition,
$
   = 1 - {\left( {\dfrac{3}{4}} \right)^n} \geqslant \dfrac{9}{{10}} \\
   = 1 - \dfrac{9}{{10}} \geqslant {\left( {\dfrac{3}{4}} \right)^n} \\
   = \dfrac{1}{{10}} \geqslant {\left( {\dfrac{3}{4}} \right)^n} \\
  {\text{Taking log both the sides in order to find n}} \\
  {\text{ = log}}\left( {\dfrac{1}{{10}}} \right) \geqslant \log \left( {{{\left( {\dfrac{3}{4}} \right)}^n}} \right) \\
   = - 1 \geqslant n\left[ {\log \left( 3 \right) - \log \left( 4 \right)} \right] \\
  {\text{multiplying by - 1 on both sides}} \\
  {\text{ = 1}} \leqslant n\left[ {\log \left( 4 \right) - \log \left( 3 \right)} \right] \\
   = \dfrac{1}{{\left[ {\log \left( 4 \right) - \log \left( 3 \right)} \right]}} \leqslant n \\
$
So, the correct option is A.

Note: Instead of P(at least one), we took [1-P(none)] since it is easy to calculate the probability of 0 success instead of counting the possibilities of success in these cases.