7.3 Mutually independent events

Three events are said to be mutually independent if they satisfy all the following four conditions:

i. Mutually exclusiveness is used when the events are taken from the same experiment and independence is used when the events are taken from different experiments.

ii. In conditional probability the sample space is reduced to the set of samples or outcomes in the event which is said to have happened.

ii. The sample space is

$$S = \{ (H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)\} $$

$A = \{ (H,1),(H,2),(H,3),(H,4),(H,5),(H,6)\} $

$B = \{ (H,3),(T,3)\} $

$$P(A) = \;{6 \over {12}} = {1 \over 2}$$

$$P(B) = \;{2 \over {12}} = {1 \over 6}$$

$$\begin{equation} \begin{aligned} n(A \cap B) = 1 \\ \Rightarrow P(A \cap B) = {1 \over {12}} \\\end{aligned} \end{equation} $$

Clearly,

$$P(A \cap B) = P(A)P(B)$$

Hence $A$ and $B$ are independent events.

iii. Total number of outcomes is $36$.

$n(A) = 2$

$n(B) = 30$.

${A \cap B}$ = $\{ (6,5)\} $.

Thus, $n(A \cap B) = 1$

$$P(A) = \;{2 \over {36}} = {1 \over {18}}$$

$$P(B) = \;{{30} \over {36}} = {5 \over 6}$$

$$P(A \cap B) = {1 \over {36}}$$

Clearly,

$$P(A \cap B) \ne P(A)P(B)$$

Thus the events are non-independent.

Question 20. A class consists of $80$ students out of which $25$ are girls and the rest are boys. $10$ of them hold prefect positions, and the rest do not. $20$ of them are sports person. Find the probability of choosing a sports girl who holds a prefect position.

Solution: Let us consider the following events,

$A$ = Selecting a sports student.

Thus,

$$P(A) = \;{{20} \over {80}} = {1 \over 4}$$

$B$ = Selecting a student holding a prefect position.

Thus,

$$P(B) = \;{{10} \over {80}} = {1 \over 8}$$

$C$ = Selecting a girl.

Thus,

$$P(C) = \;{{25} \over {80}} = {5 \over {16}}$$

Required probability is

$$\begin{equation} \begin{aligned} P(A \cap B \cap C) = P(A)P(B)P(C) \\ = {1 \over 4} \times {1 \over 8} \times {5 \over {16}} = {5 \over {512}} \\\end{aligned} \end{equation} $$

Question 21. A problem is given to $4$ students whose chances of solving it are ${1 \over 4}$, ${1 \over 3}$, ${1 \over 5}$, ${1 \over 2}$. What is the probability the problem is solved.

Solution: The events are independent. The problem is solved if at least one solves it.

Let $A$, $B$, $C$ and $D$ be the students solving the problem.

Required probability is,

$$\begin{equation} \begin{aligned} P(A \cup B \cup C \cup D) \\ = 1 - P(\bar A)P(\bar B)P(\bar C)P(\bar D) \\ = 1 - [1 - P(A)][1 - P(B)][1 - P(C)][1 - P(D)] \\ = 1 - [1 - {1 \over 4}][1 - {1 \over 3}][1 - {1 \over 5}][1 - {1 \over 2}] \\ = 1 - \left[ {{3 \over 4}} \right]\left[ {{2 \over 3}} \right]\left[ {{4 \over 5}} \right]\left[ {{1 \over 2}} \right] \\ = 1 - {1 \over 5} = {4 \over 5} \\\end{aligned} \end{equation} $$

Question 22. An urn contains four tickets with numbers $112$, $121$, $211$, $222$ and one ticket is drawn. Let ${A_i}$ $(i = 1,2,3)$ be the event the ${i^{th}}$ digit of the number on the ticket drawn is $1$. Discuss the independence of the events ${A_1},{A_2},{A_3}$.

Solution: $P({A_1})$ = Probability that the first digit of the number on the drawn ticket is $1$.

$$P({A_1}) = {2 \over 4} = {1 \over 2}$$

$P({A_2})$ = Probability that the second digit of the number on the drawn ticket is $1$.

$$P({A_2}) = {2 \over 4} = {1 \over 2}$$

$P({A_3})$ = Probability that the third digit of the number on the drawn ticket is $1$.

$$P({A_3}) = {2 \over 4} = {1 \over 2}$$

$P({A_1} \cap {A_2})$ = Probability that the first and second digits of the number on the drawn ticket are each equal to $1$

$$P({A_1} \cap {A_2}) = {1 \over 4}$$

Similarly,

$$P({A_2} \cap {A_3}) = {1 \over 4}$$

$$P({A_1} \cap {A_3}) = {1 \over 4}$$

$$P({A_1} \cap {A_2} \cap {A_3}) = 0$$

$$\begin{equation} \begin{aligned} P({A_1} \cap {A_2}) = P({A_1})P({A_2}) \\ P({A_2} \cap {A_3}) = P({A_2})P({A_3}) \\ P({A_3} \cap {A_1}) = P({A_3})P({A_1}) \\\end{aligned} \end{equation} $$

and $$P({A_1} \cap {A_2} \cap {A_3}) \ne P({A_1})P({A_2})P({A_3})$$

Thus the events are pairwise independent but not mutually independent.

Question 23. A village has two independently functioning fire engines. The probability of availability of each engine, when needed, is $0.95$. What is the probability that

i. neither of them is available when needed.

ii. an engine is available

iii. exactly one engine is available

Solution: Let $A$ be the event that first fire engine is available and $B$ be the event where second one is available. $P(A) = P(B) = \;0.95$

i. Neither of them available

$$\begin{equation} \begin{aligned} P(\bar A \cap \bar B) \\ = P(\bar A)P(\bar B) \\ = 0.05 \times 0.05 \\ = 0.0025 \\\end{aligned} \end{equation} $$

ii. At least one of them available

$$\begin{equation} \begin{aligned} P(A \cup B) \\ = 1 - P(\bar A)P(\bar B) \\ = 1 - 0.05 \times 0.05 \\ = 1 - 0.0025 \\ = 0.9975 \\\end{aligned} \end{equation} $$

iii. Exactly one of them available

$$\begin{equation} \begin{aligned} P(A) + P(B) - 2P(A \cap B) \\ = P(A) + P(B) - 2[P(A)P(B)] \\ = 0.95 + 0.95 - 2 \times 0.95 \times 0.95 \\ = 0.095 \\\end{aligned} \end{equation} $$

Question 24. A combination lock on a suitcase has $3$ wheels each labeled with nine digits from $1$ to $9$. If an opening combination is a particular sequence of three digits with no repeats, what is the probability of a person guessing the right combination?

Solution: Let ${A_i},\;i = 1,2,3$ be the event that the digit on ${i^{th}}$ wheel occupies the correct position. Then, the probability of getting the correct one is

$$\begin{equation} \begin{aligned} P({A_1} \cap {A_2} \cap {A_3}) \\ = P({A_1})P({A_2}/{A_1})P({A_3}/{A_1} \cap {A_2}) \\ = {1 \over 9} \times {1 \over 8} \times {1 \over 7} \\ = {1 \over {504}} \\\end{aligned} \end{equation} $$

Question 25. The odds against $A$ solving a problem is $4$ to $3$ and the odds in favor of $B$ solving the same problem is $7$ to $5$. Find the probability that the problem is solved.

Solution: Odds against $A$ are $4$ to $3$.

$$P(A) = {3 \over {3 + 4}} = {3 \over 7}$$

Odds in favor $B$ are $7$ to $5$.

$$P(B) = {7 \over {7 + 5}} = {7 \over {12}}$$

The problem is solved, if at least one of them solves.

Thus probability the problem is solved is

$$\begin{equation} \begin{aligned} P(A \cup B) \\ = 1 - P(\bar A)P(\bar B) \\ = 1 - \left[ {1 - {3 \over 7}} \right]\left[ {1 - {7 \over {12}}} \right] \\ = 1 - \left[ {{4 \over 7}} \right]\left[ {{5 \over {12}}} \right] \\ = 1 - \left[ {{5 \over {21}}} \right] \\ = {{16} \over {21}} \\\end{aligned} \end{equation} $$

Question 26. A number is chosen at random from the numbers $10$ to $99$. By seeing the number, a man will show a green card if the product of the digits is $18$. If he chooses three numbers with replacement then find the probability that he will show green card at least once.

Solution: For product of digits to be $18$, possible numbers are = $29$, $36$, $63$, $92$.

Total number of numbers

$$99 - 10 + 1 = 90$$

The probability that the number chosen is any of the above is

$$ = {4 \over {90}} = {2 \over {45}}$$

$$\begin{equation} \begin{aligned} The\;probability\;that\;he\;shows\;green\;card\;at\;least\;once\;is \\ = 1 - probability\;that\;he\;never\;shows \\ = 1 - \left[ {1 - {2 \over {45}}} \right]\left[ {1 - {2 \over {45}}} \right]\left[ {1 - {2 \over {45}}} \right] \\ Since,three\;times\;a\;number\;is\;chosen\;with\;replacement \\ = 1 - {\left[ {{{43} \over {45}}} \right]^3} \\\end{aligned} \end{equation} $$