Question 1198328
 Let {{{N }}}be the number of heads among the {{{n}}} tosses. Then, {{{N}}} ∼ {{{B}}}({{{n}}},{{{ p}}}). 

given:

probability of obtaining a head equal to:{{{1/3000000000=1/(3 *10^9)}}} 

You toss this coin {{{6000000000= 6*10^9}}} times


Here, we have small {{{p = 1/(3 *10^9)}}} and large {{{n = 6*10^9}}}. So, we can apply Poisson approximation.

In other words, {{{B}}}({{{n}}},{{{ p}}}) is well-approximated by {{{P(alpha)}}} where {{{alpha = np = 2}}}.


(a) 

{{{P(A)}}} = {{{P}}} [{{{N = 0}}}] = {{{e^-2(2^0/0!)=1/e^2}}}≈ {{{0.1353}}}

(b) 

{{{P}}}({{{A}}} ∪ {{{B}}})= {{{P}}} [{{{N = 0}}}]+{{{P}}} [{{{N = n}}}] = {{{e^-2(2^0/0!)+e^-2((2^6*10^9)/(6*10^9)!)}}}

 The second term is {{{extremely}}}{{{ small}}} compared to the first one. 

Hence, {{{P}}}({{{A}}} ∪ {{{B}}}) is approximately the same as {{{P(A)}}}.