Question

If the probability is 0:75 that any person will believe a rumour, nd the probabilities

that (i) the fth person to hear it is the rst to believe it (ii) the eighth person to hear

the rumour will be the fth to believe it (iii) at least 4 persons do not believe the rumour

before the tenth person believes it.

Answer 1

(i) Let $A_i$ denote the event '$i$ -th person will believe a rumour ' .

Then the required probability is $P(\bar A_1\bar A_2\bar A_3\bar A_4A_5)$

Here $P(A_i)=0.75=\frac{3}{4}$ and $P(\bar A_i)=0.25=\frac{1}{4}$

$\therefore$ $P(\bar A_1\bar A_2\bar A_3\bar A_4A_5)=P(\bar A_1).P(\bar A_2).P(\bar A_3).P(\bar A_4).P( A_5)$ [ as all the events are independent]

$=\frac{1}{4}.\frac{1}{4}.\frac{1}{4}.\frac{1}{4}.\frac{3}{4}=\frac{3}{1024}$

(ii) According to the problem eighth person to hear the rumour will the the fifth to believe. Therefore upto seventh person ,those who hear the rumour only four of them will believe the rumour and eighth person will believe the rumour.

Let ' success' denote the event ' person will believe the rumour'

Then $p=$ probability of success $=\frac{3}{4}$

$q=$ probability of failure $=\frac{1}{4}$

Then probability of four success out of seven trial is $= {^7C_4}.p^4.q^{(7-4)}=35.(\frac{3}{4})^4.(\frac{1}{4})^3=35.\frac{3^4}{4^7}$

So our required probability that eighth person to hear the rumour will be the fifth to believe is = $35.(\frac{3^4}{4^7}).(\frac{3}{4})$ [as eighth person believe the rumour]

$\therefore$ Required probability $=35.(\frac {3^4}{4^7}).(\frac{3}{4})=35.\frac{3^5}{4^8}$

(iii) We have to find at least $4$ persons don't believe the rumour before the tenth person believe, i.e at least 4 person don't believe the rumour of $9$ persons.

First we calculate at most $3$ person don't believe the rumour.

Let ' success' denote the event ' person will believe the rumour'

Then $p=$ probability of success $=\frac{3}{4}$

$q=$ probability of failure $=\frac{1}{4}$

Also let $X$ be random variable defined as $X=$ number of person don't believe the rumour

Then $X$ can take the values $0,1,2,3$

Now $P(X=0)={^9C_0}.p^0.q^{(9-0)}=(\frac{1}{4})^9$

$P(X=1)={^9C_1}.p^1.q^{(9-1)}=9.(\frac{3}{4}).(\frac{1}{4})^8=\frac{27}{4^9}$

$P(X=2)={^9C_3}.p^2.q^{(9-2)}=36.(\frac{3}{4}).^2(\frac{1}{4})^7=\frac{324}{4^9}$

$P(X=3)={^9C_3}.p^3.q^{(9-3)}=84.(\frac{3}{4})^3.(\frac{1}{4})^6=\frac{2268}{4^9}$

So probability of at least $4$ persons do not believe the rumour is $=1-(\frac{1}{4^9}+\frac{27}{4^9}+\frac{324}{4^9}+\frac{2268}{4^9})$

$=1-(\frac{2620}{4^9})$

$\therefore$ The required probability that at least $4$ persons do not believe the rumour before the tenth person believe is

$=[1-(\frac{2620}{4^9})]×(\frac{3}{4})=\frac{3}{4}-\frac{7860}{4^{10}}$