91Ó°ÊÓ

The ensuing data defines the conditional probabilities of obtaining the message of mail received or rejected on each day of next week:

The probability of being received is,$P\left(A\right)=0.6$

a)

The purpose is to compute that she obtains mail on Monday.

$P\left(\text{Receive mail on Monday}\right)=\left(\begin{array}{l}P\left(\text{Monday Accepted}\right)P\left(\text{Accepted}\right)+\\ P\left(\text{Monday Rejected}\right)(1−P(\text{Accepted}\left)\right)\end{array}\right)$

$=(0.15×0.6)+\left(0.05\right(1−0.6\left)\right)$

$=0.09+0.02$

$=0.11$

Therefore, the required probability is,$0.11$.

b)

The purpose is to compute the conditional probability that she got mail on Tuesday given that she does not acquire mail on Monday.

$\begin{array}{r}P\left(\text{Tuesday}âˆ\pounds (\text{Monday}{)}^c\right)=\frac{\left(\begin{array}{c}P\left(\text{Tuesday Accepetd}\right)P\left(\text{Accepetd}\right)\\ +P\left(\text{Tuesday Rejected}\right)P\left(\text{Rejected}\right)\end{array}\right)}{P(\text{Monday}{)}^c}\\ \\ \end{array}$ localid="1646999238587" $=\frac{(0.20×0.6)+\left(0.10\right(1−0.6\left)\right)}{1−P\left(\text{Monday}\right)}$

$=\frac{0.12+0.04}{1−0.11}$

localid="1646999335365" role="math" $\begin{array}{r}=\frac{0.16}{0.89}\end{array}$

$=0.1798$

Hence, the required probability is $0.1798$

c)

The purpose is to discover the conditional probability that she will be accepted if there is no mail via Wednesday.

Let $A\text{and}R$represent the Accepted and Rejected.

$P\left(\begin{array}{l}\text{Accepted if}\\ \text{there is no}\\ \text{mail through}\\ \text{Wednessday}\end{array}\right)=\frac{P\left(\text{A}\right)\left(\begin{array}{l}1−P\left(\text{Mon Aceepted}\right)+\\ P\left(\text{Tue Aceepted}\right)+\\ P\left(\text{Wed Aceepted}\right)\end{array}\right)}{P\left(\text{A}\right)\left(\begin{array}{l}1−P\left(\text{Mon Aceepted}\right)+\\ P\left(\text{Tue Aceepted}\right)+\\ P\left(\text{Wed Aceepted}\right)\end{array}\right)+P\left(\mathrm{R}\right)\left(\begin{array}{l}1−P\left(\text{Mon Rejected}\right)+\\ P\left(\text{Tue Rejected}\right)+\\ P\left(\text{Wed Rejected}\right)\end{array}\right)}$

$=\frac{0.6(1−(0.15+0.20+0.25\left)\right)}{0.6(1−0.15+0.20+0.25)+(1−0.6)(1−0.05+0.10+0.10)}$

$=\frac{0.6(1−0.6)}{0.6(1−0.6)+0.4(1−0.25)}$

$=\frac{0.6×0.4}{(0.6×0.4)+(0.4×0.75)}$

$=\frac{0.24}{0.24+0.30}$

$\begin{array}{r}=\frac{0.24}{0.54}\end{array}$

$=0.4444444$

$=0.4444$

Therefore the required probability is,$0.4444\text{.}$

d)

The purpose is to compute the conditional probability that she will be accepted if mail comes on Thursday.

$P\left(\begin{array}{l}\text{Accpted mail}\\ \text{on Thursday}\end{array}\right)=\frac{P\left(\text{Accepted}\right)P\left(\text{Thursday Accepted}\right)}{\left(\begin{array}{l}P\left(\text{Accepted}\right)P\left(\text{Thursday Accepted}\right)+\\ P\left(\text{Rejected}\right)P\left(\text{Thursday Rejected}\right)\end{array}\right)}$

$=\frac{0.6×0.15}{(0.6×0.15)+(1−0.6)\left(0.15\right)}$

$=\frac{0.09}{0.09+0.06}$

$=\frac{0.09}{0.15}$

$=0.6$

Therefore, the required probability is$0.60\text{.}$

e)

The purpose is to compute the conditional probability that she will be accepted if no mail reaches that week.

$P\left(\text{Accpted no mail}\right)=\frac{P\left(\text{A}\right)P\left(\text{No mail accepted}\right)}{\left(\begin{array}{l}P\left(\text{No mail accepted}\right)P\left(A\right)+\\ P\left(\text{No mail rejected}\right)P\left(R\right)\end{array}\right)}$

$=\frac{0.6(1−(0.15+0.20+0.25+0.15+0.10\left)\right)}{\left(\begin{array}{l}(1−(0.15+0.20+0.25+0.15+0.10\left)\right)0.6+\\ (1−0.05+0.10+0.10+0.15+0.20)(1−0.6)\end{array}\right)}$

$=\frac{0.6(1−0.85)}{(1−0.85)0.6+(1−0.6)0.4}$

$=\frac{0.6×0.15}{0.09+0.16}$

localid="1646999404252" $\begin{array}{rr}=\frac{0.09}{0.25}& \end{array}$

$=0.36$

Therefore, the required probability is $0.36$