Discussion Forum

Let X be the number of successes in 'n'  independent Bernoulli trials with probability of success p= $\frac{3}{4}$. The least value of 'n'  so that   $P(X\geq1)\geq 0.9375$  is 

Answer:

Explanation:

We have ,  p= $\frac{3}{4}$ q=1-p = $\frac {1}{4}$

 It is given that $P(X\geq1)\geq 0.9375$ 

= $1-P(X=0)\geq 0.9375$

 = $1-^{n}C_{0}(p^{0})(g)^{n-0}\geq 0.9375$

 = $1-\left(\frac{1}{4}\right)^{n}\geq 0.9375$

  = $1-0.9375\geq\left(\frac{1}{4}\right)^{n}$

   =$0.0625\geq\left(\frac{1}{4}\right)^{n}$

 = $\frac{625}{10000}\geq\left(\frac{1}{4}\right)^{n}$

 =$\frac{1}{16}\geq\left(\frac{1}{4}\right)^{n}$

 =  ${16}\leq4^{n}$

= n=2