Question 1205089


 

given:

{{{P(X<=3)}}}
{{{n=7}}}
{{{p=0.6}}}


At most {{{3 }}}successes includes {{{X}}}-values of {{{X}}}={{{0}}},{{{1}}},{{{2}}},{{{3}}}

To solve this problem, find the sum of the binomial probabilities for each of the values of {{{X}}}, or if there is only one value of {{{X}}}, find the probability of {{{P(X)}}}. 

In this problem,{{{ P(0)+P(1)+P(2)+P(3)}}}.


{{{P(0)=7!/(0!(7-0)!)*0.6^0*(1-0.6)^(7-0)=0.0016384}}}
{{{P(1)=7!/(1!(7-1)!)*0.6^1*(1-0.6)^(7-1)=0.0172032}}}
{{{P(2)=7!/(2!(7-2)!)*0.6^2*(1-0.6)^(7-2)=0.0774144}}}
{{{P(3)=7!/(3!(7-3)!)*0.6^3*(1-0.6)^(7-3)=0.193536}}}


{{{P(0)+P(1)+P(2)+P(3)=0.0016384+0.0172032+0.0774144+0.193536=0.2898}}}

answer: {{{P(X<=3)=0.2898}}}