Question 694369
1.

Probability of event {{{A}}} that occurs:

 {{{P(A) = n(A) / n(S)}}}

 {{{P(A) = 4 / 5}}}

 {{{P(A) = highlight(0.8)}}}

Probability of event {{{A}}} that does not occur:

 {{{P}}}({{{A}}}') = {{{1 - P(A)}}} 

{{{P}}}({{{A}}}') = {{{1 -  0.8}}}

{{{P}}}({{{A}}}')= {{{0.2}}}


2.

A copy machine randomly puts out 10 blank sheets per 500 copies processed. 
 {{{p=10/500=highlight(0.02)}}}

the probability that in a run of {{{300 }}}copies, {{{5 }}}sheets of paper will be blank is

{{{p=5/300=0.016666666666666666666666666666667}}} ≈ {{{highlight(p=0.017)}}}

3.

{{{Prob( a robbery) = 20000/80000=1/4 = 0.25}}}

{{{n=80000}}}

The Binomial distribution has the probability

function {{{P(x=r)= nCr p^r (1-p)^(n-r)}}}

{{{r}}}={{{0}}},{{{1}}},{{{2}}},.....,{{{n}}}

where {{{nCr = n! / (r! (n-r)!)}}}

For {{{80000}}} people, we expect {{{4}}} robberies.

a)
{{{n=80000}}}

{{{p=4/80000}}}

{{{p=1/20000}}}

{{{highlight(r=0.00005)}}}



use the source to compute

{{{(80000C0)(1/20000)^0(19999/20000)^8000= 0.018}}}

Using Poisson with {{{lambda =4}}}

{{{P(x=0)= e^-4 *(4^0 / 1!)=0.018(1/1)=highlight(0.018)}}}


b)


{{{highlight(r=1)}}}

Using Poisson with lambda=4

{{{e^-4(4^1 / 1!)=0.018*4=highlight(0.072)}}}


c)

{{{highlight(r=3)}}}


{{{e^-4 (4^3 / 3!) =0.018*(64/6)=0.018*10.666666666666666666666666666667= highlight(0.192)}}}

d) 

p( highlight(x >=3))= p( 3 through 20000)

{{{np=80000(0.00005)=4}}}

{{{variance=np(1-p)=3.99998}}}

{{{sd=1.99999}}}

{{{P( x >=3) = P( z >(3-4)/2) = p( z > -0.5)=highlight(0.6915)}}}

This uses normal approximation instead of Binomial because {{{n}}} is too large and {{{p}}} is too small.

4.

Presuming there are {{{200}}} of {{{400}}} total pages with misprints, there is a {{{50}}}% probability of discovering one on any given page.

{{{p=200/400=1/2=0.50}}} which is {{{highlight(50)}}}%