Q.3.52 - A First Course in Probability

Question

A high school student is anxiously waiting to receive mail telling her whether she has been accepted to a certain college. She estimates that the conditional probabilities of receiving notification on each day of next week, given that she is accepted and that she is rejected, are as follows: 

Day	P(mail/accepted)	P(mail/rejected)
Monday	$.15$	$.05$
Tuesday	$.20$	$.10$
Wednesday	$.25$	$.10$
Thursday	$.15$	$.15$
Friday	$.10$	$.20$

She estimates that her probability of being accepted is .6. 

(a) What is the probability that she receives mail on Monday? 

(b) What is the conditional probability that she receives mail on Tuesday given that she does not receive mail on Monday? 

(c) If there is no mail through Wednesday, what is the conditional probability that she will be accepted? 

(d) What is the conditional probability that she will be accepted if mail comes on Thursday? 

(e) What is the conditional probability that she will be accepted if no mail arrives that week? 

Step-by-Step Solution

Answer

a)$0.11$ is the probability that she receives mail on Monday.

b)$0.1798$ is the conditional probability that she receives mail on Tuesday given that she does not receive mail on Monday.

c)$0.4444$ is the conditional probability that she will be accepted if there is no mail through Wednesday.

d)$0.60$ is the conditional probability that she will be accepted if mail comes on Thursday.

e)$0.36$ is the conditional probability that she will be accepted if no mail arrives that week.

Step 1:Given Information

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

Day	P(mail/accepted)	P(mail/rejected)
Monday	$0.15$	$0.05$
Tuesday	$0.20$	$0.10$
Wednesday	$0.25$	$0.10$
Thursday	$0.15$	$0.15$
Friday	$0.10$	$0.20$

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

Step 2:Explanation and Final answer(Part a)

a)

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

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

$=(0.15\times 0.6)+\left(0.05\right(1-0.6\left)\right)$

$=0.09+0.02$

$=0.11$

Therefore, the required probability is,$0.11$.

Step 3:Explanation and Final Answer(Part b)

b)

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

$\begin{array}{r}P\left(\text{ Tuesday }\mid (\text{ Monday }{)}^c\right)=\frac{\left(\begin{array}{c}P\left(\text{ Tuesday Accepetd }\right)P\left(\text{ Accepetd }\right)\\ +P\left(\text{ Tuesday Rejected }\right)P\left(\text{ Rejected }\right)\end{array}\right)}{P(\text{ Monday }{)}^c}\\ \\ \end{array}$    $=\frac{(0.20\times 0.6)+\left(0.10\right(1-0.6\left)\right)}{1-P\left(\text{ Monday }\right)}$

$=\frac{0.12+0.04}{1-0.11}$

$\begin{array}{r}=\frac{0.16}{0.89}\end{array}$

$=0.17977$

$=0.1798$

Hence, the required probability is $0.1798$

Step 4:Explanation and Final Answer(Part c)

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.

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

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

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

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

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

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

$=0.4444444$

$=0.4444$

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

Step 5:Explanation and Final Answer(Part d)

d)

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

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

$=\frac{0.6\times 0.15}{(0.6\times 0.15)+(1-0.6)\left(0.15\right)}$

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

$=\frac{0.09}{0.15}$

$=0.6$

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

Step 6:Explanation and Final Answer(Part e)

e)

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

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

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

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

$=\frac{0.6\times 0.15}{0.09+0.16}$

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

$=0.36$

Therefore, the required probability is $0.36$