Probability
8.0 Total Probability Theorem
8.0 Total Probability Theorem
Let ${A_1},{A_2},....,{A_n}$ be a set of mutually exclusive and exhaustive events and $B$ be some event which is associated with ${A_1},{A_2},....,{A_n}$. Then probability that $B$ occurs is given by $$P(B) = \sum\limits_{i = 1}^n {P({A_i})} P(B/{A_i})$$
Illustration 35. Find the probability that a year chosen at random has $53$ Sundays.
Solution: Let $P(L)$ be the probability that a year chosen at random is a leap year.
$$\begin{equation} \begin{aligned} P(L) = {1 \over 4} \\ P(\bar L) = {3 \over 4} \\\end{aligned} \end{equation} $$
Let $P(S)$ be the probability that an year chosen at random has $53$ Sundays
Thus,
$$P(S) = P(\bar L)P(S/\bar L) + P(L)P(S/L)$$
Where $P(S/\bar L)$ is the probability that there are $53$ Sundays in a non-leap year.
There are $52$ weeks and one day. Thus the chance of the extra one day being Sunday is $$ = {1 \over 7}$$
and,
$P(S/L)$ is the probability that there are $53$ Sundays in a leap year.
There are $52$ weeks and two days. This can be Mon-Tue, Tue-Wed, Wed-Thur, Thur-Fri, Fri-Sat, Sat-Sun, Sun-Mon. Thus the chance of being Sunday is $$ = {2 \over 7}$$
$$\begin{equation} \begin{aligned} P(S) = P(\bar L)P(S/\bar L) + P(L)P(S/L) \\ P(S) = {3 \over 4} \times {1 \over 7} + {1 \over 4} \times {2 \over 7} = {5 \over {28}} \\\end{aligned} \end{equation} $$
Illustration 36. One bag contains $4$ white and $5$ black balls. Another bag contains $6$ white and $7$ black balls. A ball is transferred from first bag to the second bag and then a ball is drawn from the second bag. Find the probability that the ball so drawn is white.
Solution: A white ball can be drawn from the second bag by two ways.
By transferring a white ball from first bag to the second, and then drawing a ball or by transferring a black ball from first bag to the second, and then drawing a ball.
Let $W$ be an event that a white ball is transferred and $B$ an event that black ball is transferred.
Let $D$ be the event of a white ball being drawn.
$$\begin{equation} \begin{aligned} P(W) = {4 \over 9} \\ P(B) = {5 \over 9} \\\end{aligned} \end{equation} $$
When $W$ occurs, the second bag will have $7$ black and $7$ white balls, and when $B$ occurs it will have $6$ white and $8$ black balls.
Thus,
$$\begin{equation} \begin{aligned} P(D/W) = {7 \over {14}} \\ P(D/B) = {6 \over {14}} \\\end{aligned} \end{equation} $$
Therefore,
$$\begin{equation} \begin{aligned} P(D) = P(W)P(D/W) + P(B)P(D/B) \\ P(D) = {4 \over 9} \times {7 \over {14}} + {5 \over 9} \times {6 \over {14}} \\ P(D) = {{58} \over {126}} \\ P(D) = {{29} \over {63}} \\\end{aligned} \end{equation} $$
Question 27. There are two bags. The first bag contains $5$ white and $3$ black balls and the second bag contains $3$ white and $5$ black ball. Two balls are drawn from the first and are put into the second bag without noticing their colors. Then two balls are drawn from the second bag. Find the probability that the balls are white and black.
Solution: A white and a black ball can be drawn from the second bag in three ways.
By transferring two black balls and then drawing a black and a white ball, or by transferring two white balls and then drawing a black and a white ball, or by transferring one black and one white ball and then drawing a black and a white ball.
$A$ is an event that two black balls are transferred, $B$ is an event that two white balls are transferred and $C$ is an event that one black and one white ball is transferred.
Let $D$ be the event that two balls drawn from the second bag are white and black.
$$\begin{equation} \begin{aligned} P(A) = {{^3{C_2}} \over {^8{C_2}}} = {3 \over {28}} \\ P(B) = {{^5{C_2}} \over {^8{C_2}}} = {5 \over {14}} \\ P(C) = {{^3{C_1}{ \times ^5}{C_1}} \over {^8{C_2}}} = {{15} \over {28}} \\\end{aligned} \end{equation} $$
When $A$ happens, there will be $3$ white and $7$ black balls.
Thus,
$$P(D/A) = {{^3{C_1}{ \times ^7}{C_1}} \over {^{10}{C_2}}} = {7 \over {15}}$$
Similarly, when $B$ happens, there will be $5$ white and $5$ black balls.
Thus,
$$P(D/B) = {{^5{C_1}{ \times ^5}{C_1}} \over {^{10}{C_2}}} = {5 \over 9}$$
and when $C$ happens, there will be $4$ white and $6$ black balls.
Thus,
$$P(D/C) = {{^4{C_1}{ \times ^6}{C_1}} \over {^{10}{C_2}}} = {8 \over {15}}$$
Therefore,
$$\begin{equation} \begin{aligned} P(D) = P(A)P(D/A) + P(B)P(D/B) + P(C)P(D/C) \\ P(D) = {3 \over {28}} \times {7 \over {15}} + {5 \over 9} \times {5 \over {14}} + {{15} \over {28}} \times {8 \over {15}} \\ P(D) = {{673} \over {1260}} \\\end{aligned} \end{equation} $$
Question 28. A person has undertaken a construction job. The probabilities are $0.65$ that there will be strike, $0.80$ that the construction job will be completed on time if there is no strike, and $0.32$ that the construction job will be completed on time if there is strike. Determine the probability that the construction job will be completed on time.
Solution: Let $A$ be the event that the construction job will be completed on time.
And, let ${E_1}$ be the event that there will be a strike, and ${E_2}$, that there will be no strike.
Probability that there will be strike $$P({E_1}) = 0.65$$
Probability that there will not be strike $$P({E_2}) = 0.35$$
Probability that the work will be completed, given there is no strike is $$P\left( {A/{E_1}} \right) = 0.80$$
Probability that the work will be completed given that there was a strike is $$P\left( {A/{E_2}} \right) = 0.32$$
Thus the probability that the work will be completed is $P(A)$.
$$\begin{equation} \begin{aligned} P(A) = P({E_1})P(A/{E_1}) + P({E_2})P(A/{E_2}) \\ P(A) = 0.65 \times 0.32 + 0.35 \times 0.80 \\ P(A) = 0.208 + 0.28 \\ P(A) = 0.488 \\\end{aligned} \end{equation} $$
Thus the probability that the construction will be complete on time is $0.488$.
