3.2 Odds in favor

  Probability

If out of '$m + n$' equally likely, mutually exclusive and exhaustive events, '$m$' are favourable to occurrence of an event $A$ and '$n$' are not favourable, then $m:n$ is called odds in favor of $A$, and $n:m$ is called odds against $A$.

Illustration 10. If the probability of Shyam winning a game is ${5 \over {13}}$ and Kiran losing the game is ${4 \over {12}}$, express each of their probabilities as odds in favor and odds against each of them.

Solution: Let us consider Shyam,

$$\begin{equation} \begin{aligned} P(winning) \\ = {5 \over {13}} \\ = {5 \over {5 + 8}} \\ = {m \over {m + n}} \\ \Rightarrow m = 5\;and\;n = 8 \\\end{aligned} \end{equation} $$

Thus odds in favor of Shyam is $5:8$ or in other words, $5$ to $8$.

Thus odds against of Shyam is $8:5$ or in other words, $8$ to $5$.

Let us consider Kiran,

$$\begin{equation} \begin{aligned} P(losing) \\ = {4 \over {12}} \\ = {4 \over {4 + 8}} \\ = {n \over {n + m}} \\ \Rightarrow m = 8\;and\;n = 4 \\\end{aligned} \end{equation} $$

Thus odds in favor of Kiran is $8:4$ or in other words, $8$ to $4$.

Thus odds against of Kiran is $4:8$ or in other words, $4$ to $8$.

Illustration 11. There are $3$ mutually exclusive and exhaustive events, $A$ , $B$ and $C$. Odds in favor of $A$ is $5$ to $6$ and probability of success of $B$ is ${3 \over {11}}$. Find the success probability of $C$. Express the same as odds in favor.

Solution: Given odds in favor of $A$ is $5:6$.

Thus the probability is

$$\begin{equation} \begin{aligned} = {5 \over {5 + 6}} \\ = {5 \over {11}} \\\end{aligned} \end{equation} $$

And the probability of $B$ is ${3 \over {11}}$

Since the events $A$, $B$ and $C$ are mutually exclusive and exhaustive events,

$$P(A) + P(B) + P(C) = 1$$

Therefore

$$\begin{equation} \begin{aligned} P(C) = 1 - P(A) - P(B) \\ P(C) = 1 - {5 \over {11}} - {3 \over {11}} \\ P(C) = {3 \over {11}} \\\end{aligned} \end{equation} $$

Thus the probability of success is ${3 \over {11}}$. The odds in favor is $3:8$ or $3$ to $8$.

Question 1. A die is thrown twice. Find the probability of following events.

i. A doublet

ii. At least one five

iii. Sum is seven.

Solution: There are $6 \times 6$ equally likely cases. Thus $36$ total possible outcomes.

i. A doublet $ \Rightarrow \{ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\} $

Thus probability of doublet is $ = {6 \over {36}} = {1 \over 6}$

ii. At least one five $ \Rightarrow \{ (1,5),(2,5),(3,5),(4,5),(5,5),(6,5),(5,1),(5,2),(5,3),(5,4),(5,6)\} $

Thus probability of getting at least one $5$ is $ = {{11} \over {36}}$

iii. The sum is seven $ \Rightarrow \{ (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\} $

Thus probability that the sum is seven is $ = {6 \over {36}} = {1 \over 6}$

Question 2. From a pack of $52$ cards, two cards are drawn. Find the probability that

i. They are of the same flower type

ii. One hearts and another diamond.

Solution: The total outcomes possible is

$$\begin{equation} \begin{aligned} ^{52}{C_2} \\ = {{52 \times 51} \over {1 \times 2}} \\ = 1326 \\\end{aligned} \end{equation} $$

i. They are of same flower. Let the flower be clubs.

Thus the number of elements in $A$ is,

$$\begin{equation} \begin{aligned} ^{13}{C_2} \\ = {{13 \times 12} \over {1 \times 2}} \\ = 78 \\\end{aligned} \end{equation} $$$$P(A) = {{78} \over {1326}} = {1 \over {17}}$$

ii. As one from hearts and one from diamond is considered. The number of elements in this event $B$ is,

$$\begin{equation} \begin{aligned} ^{13}{C_1}{ \times ^{13}}{C_1} \\ = 13 \times 13 \\ = 169 \\\end{aligned} \end{equation} $$$$P(B) = {{169} \over {1326}} = {{13} \over {102}}$$

Question 3. A bag of equally sized gift boxes contains $5$ green boxes, $4$ blue boxes and $7$ red boxes. Find the probability that three boxes taken at random are

i. all green

ii. all red

iii. all blue.

Solution:

The total number of boxes is $16$. Total number of ways in which $3$ boxes can be drawn is

$$\begin{equation} \begin{aligned} ^{16}{C_3} \\ = {{16 \times 15 \times 14} \over {1 \times 2 \times 3}} \\ = 560 \\\end{aligned} \end{equation} $$

i. Let the event of all three balls being green be $A$. The number of ways in which $3$ green balls can be drawn is

$$^5{C_3} = {{5 \times 4 \times 3} \over {1 \times 2 \times 3}} = 10$$

Thus $$P(A) = {{10} \over {560}} = {1 \over {56}}$$

ii. Let the event of all three balls being red be $B$. The number of ways in which $3$ red balls can be drawn is $$^7{C_3} = {{7 \times 6 \times 5} \over {1 \times 2 \times 3}} = 35$$

Thus $$P(B) = {{35} \over {560}} = {1 \over {16}}$$

iii. Let the event of all three balls being blue be $C$. The number of ways in which $3$ blue balls can be drawn is $$^4{C_3} = 4$$

Thus $$P(C) = {4 \over {560}} = {1 \over {140}}$$

Question 4. In a competition, odds in favor of student $A$ is $1:5$, against $B$ is $8:1$. Odds in favor of $D$ is $1:2$ and that against $C$ is $4:1$. Find the chances that one of them wins the competition.

Solution:

Odds in favor of $A$ $ \Rightarrow P(A) = {1 \over {1 + 5}} = {1 \over 6}$

Odds in favor of $D$ $ \Rightarrow P(D) = {1 \over {1 + 2}} = {1 \over 3}$

Odds against $B$ $ \Rightarrow P(\bar B) = {8 \over {8 + 1}} = {8 \over 9}$

Thus $$P(B) = 1 - P(\bar B) = 1 - {8 \over 9} = {1 \over 9}$$

Odds against $C$ $ \Rightarrow P(\bar C) = {4 \over {4 + 1}} = {4 \over 5}$

Thus $$P(C) = 1 - P(\bar C) = 1 - {4 \over 5} = {1 \over 5}$$

Since the above events are mutually exclusive, $$P(A \cup B \cup C \cup D) = P(A) + P(B) + P(C) + P(D)$$

$$P(A \cup B \cup C \cup D) = {1 \over 6} + {1 \over 3} + {1 \over 9} + {1 \over 5} = {{73} \over {90}}$$

Question 5. Find the probability that birthdays of five different people fall in exactly two calendar months.

Solution: Since each person can have the birth days in any of the $12$ months, each person has $12$ options. Thus total number of ways in which $5$ people can have their birthdays is $$ = 12 \times 12 \times 12 \times 12 \times 12 = {12^5}$$

Out of $12$ months, $2$ months are chosen in $^{12}{C_2}$ ways. Now birthdays of $5$ persons can fall in two months in ${2^5}$ ways. When all five birthdays fall in a month, the condition is not obeyed. There are $2$ such ways possible. Thus, there are ${2^5} - 2$ ways.

Probability that birthdays of five different people will fall in exactly two calendar months is $$ = {{^{12}{C_2} \times ({2^5} - 2)} \over {{{12}^5}}} = {{66 \times 30} \over {{{12}^5}}} = {{55} \over {(4)({{12}^3})}}$$

Question 6. Person $A$ has $2$ shares in a lottery containing $4$ prizes and $5$ blanks (losing in lottery). Person $B$ has $3$ shares in a lottery containing $3$ prizes and $6$ blanks. Compare their chances of winning.

Solution: Person $A$ has total lottery as $4 + 5 = 9$. He has two shares. Thus total outcomes possible is $^9{C_2}$. His chances of winning is possible if he gets at least one from the prizes, i.e. he does not get both blanks. Thus $$P(winning) = 1 - P(not\;winning)$$

$$n(not\;winning){ = ^5}{C_2}$$

Thus $$P(winning) = 1 - P(not\;winning)$$

$$P\left( {winning} \right) = 1 - {\rm{ }}{{^5{C_2}} \over {^9{C_2}}} = 1 - {{10} \over {36}} = 1 - {5 \over {18}} = {{13} \over {18}}$$

Person $B$ has total lottery as $3 + 6 = 9$.

He has three shares. Thus total outcomes possible is $$^9{C_3}$$

His chances of winning is possible if he gets at least one from the prizes, i.e. he does not get all three blanks.

Thus $$P(winning) = 1 - P(not\;winning)$$

$$n(not\;winning){ = ^6}{C_3}$$

Thus,

$$P(winning) = 1 - {\rm{ }}{{^6{C_3}} \over {^9{C_3}}} = 1 - {{20} \over {84}} = 1 - {{10} \over {17}} = {7 \over {17}}$$$${{P(A\;winning)} \over {P(B\;winning)}} = {{13/18} \over {7/17}} = {{221} \over {126}} \cong 1.75$$

Thus $A$ has more chances of winning than $B$.

Question 7. There are four events $A$, $B$, $C$ and $D$ out of which one and only one can happen. Odds are $5$ to $2$ against $A$, $1$ to $10$ favoring $B$ and $8$ to $3$ against $C$. Find odds against $D$.

Solution: Odds in favor of $B$ $ \Rightarrow P(B) = {1 \over {1 + 10}} = {1 \over {11}}$

Odds against $A$ $ \Rightarrow P(\bar A) = {5 \over {5 + 2}} = {5 \over 7}$

Thus $$P(A) = 1 - P(\bar A) = 1 - {5 \over 7} = {2 \over 7}$$

Odds against $C$ $ \Rightarrow P(\bar C) = {8 \over {8 + 3}} = {8 \over {11}}$

Thus $$P(C) = 1 - P(\bar C) = 1 - {8 \over {11}} = {3 \over {11}}$$

Since the above events are mutually exclusive and exhaustive, $$P(A \cup B \cup C \cup D) = P(A) + P(B) + P(C) + P(D) = 1$$$${1 \over {11}} + {2 \over 7} + {3 \over {11}} + p = 1$$$$p = 1 - \left[ {{1 \over {11}} + {2 \over 7} + {3 \over {11}}} \right] = 1 - {{50} \over {77}} = {{27} \over {77}}$$$$P(D) = {{27} \over {77}}$$ $$P(\overline D ) = {{50} \over {77}}$$

Thus odds against $D$ is $50:27$.

Question 8. Each coefficient in the equation $a{x^2} + bx + c = 0$ is determined by throwing a fair die. Find the probability that the roots of the equation are real. Also find the probability of the roots being real and equal.

Solution : Given that each of the coefficients $a$, $b$ and $c$, take the values from $1$ to $6$, i.e. all the possible outcomes of a fair die.

Since each of them can have $6$ ways, hence total possible equation depends on the total number of ways the coefficients can be filled.

Total possible ways is $6 \times 6 \times 6 = 216$.

Therefore $216$ equations can be formed.

For an equation to have real roots, the discriminant must be greater than or equal to $0$.

$$\begin{equation} \begin{aligned} Disc \ge 0 \\ {b^2} - 4ac \ge 0 \\ {b^2} \ge 4ac \\\end{aligned} \end{equation} $$

The highest value that $b$ can take is the highest value the die can show, i.e. $6$.

Thus $36$ is the highest value that ${b^2}$ can take.

Therefore, $$\begin{equation} \begin{aligned} {b^2} \ge 4ac \\ 36 \ge 4ac\; \\ ac \le 9 \\\end{aligned} \end{equation} $$

And since the values of $a$ and $c$ each depend on the outcome of a die, they cannot be $0$.

Let us now chart them into a table, considering the above derived conditions.