Chapter 16: Probability

Ans: Given: The letters of the word ALGORITHM are arranged.

Find the total number of sample spaces by finding the number of words that can be formed with the given letters of the word.

Now, find the favourable number of arrangements by considering the letters GOR always together.

There are $9$ different letters in the word ALGORITHM, so the total number of words that can be formed is the number of sample spaces $n\left( S \right) = 9!$.

Now, the letters GOR always remain together. These can be considered as a single unit, so the remaining $6$ letters and this single unit makes a total of $7$ things which can be arranged in $7!$ ways.

So the number of favourable arrangements is $n\left( E \right) = 7!$.

Hence, the required probability 

$= \dfrac{{n\left( S \right)}}{{n\left( E \right)}}$

$= \dfrac{{7!}}{{9!}}$

$= \dfrac{1}{{72}}$ 

Ans: Given: Total number of new employees $= 6$

Total number of desks in the assigned row $= 6$

A married couple is present among the $6$ employees.

Find the total number of sample space by arranging all the six employees without any condition.

Now, find the number of arrangements in which the couple always gets the adjacent seat. Subtract it from the total number of sample spaces to get the favourable sample spaces.

The $6$ new employees can be arranged at $6$ desks (without any condition) in $6!$ ways, so the number of sample space is $n\left( S \right) = 6!$.

Let us consider that the couple gets the adjacent seat, so the two people in this couple can be arranged among themselves in $2!$ ways. This couple can be considered as a single unit, so the remaining four employees and this single unit can be considered as $5$ things which can be arranged in $5!$ ways.

The total number of arrangements where the couple is always together $= 2!{\text{ }} \times {\text{ }}5!$.

Therefore, the number of arrangements where the coupe in never together $n\left( E \right) = 6!{\text{ }} - {\text{ }}\left( {2!{\text{ }} \times {\text{ }}5!} \right)$.

Hence, the required probability

$= \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$

$= \dfrac{{6!{\text{ }} - \left( {2!{\text{ }} \times {\text{ }}5!} \right)}}{{6!}}$

$= \dfrac{2}{3}$ 

Ans: Given: An integer from $1$ to $1000$ is chosen at random.

Find the total number of sample space by selecting any one integer out of a total of $1000$ integers given.

Find the numbers that are divisible by $2$ or $9$ and count the number of integers which are divisible by both to get the favourable sample spaces.

The total number of integers is from $1$ to $1000$ is thousand, so the number of ways in which one can be selected is the total number of sample space $n\left( S \right) = {}^{1000}{C_1}$.

Now, all the even integers are divisible by $2$, so there will be $500$ even numbers from $1$ to $1000$ and any one of them can be selected in ${}^{500}{C_1}$ ways. The integers that are divisible by $9$ are $9$, $18$, $27$, $36$, …, $999$. There are $111$ such integers divisible by $9$.

Among these $111$ integers there will be integers like $18$, $36$, $54$, … etc. which are also divisible by $2$ and they are already considered earlier. There are $55$ such integers. So the number of integers that is to be considered is $\left( {111 - 55} \right) = 56$ and any one of them can be chosen in ${}^{56}{C_1}$ ways.

Therefore, the number of favourable sample spaces $n\left( E \right) = {}^{500}{C_1} + {}^{56}{C_1}$.

Hence, the required probability is

$= \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$

$= \dfrac{{{}^{500}{C_1} + {}^{56}{C_1}}}{{{}^{1000}{C_1}}}$

$= 0.556$ 

Ans: Given: A die is rolled until a $2$ appears.

consider that the first $\left( {k - 1} \right)$ rolls of the die results in any of the five outcomes except $2$.

There are $6$ outcomes when a die is rolled.

In this case $2$ appears on the ${k^{th}}$ roll, so in the first $\left( {k - 1} \right)$ roll there are $5$ possible outcomes other than $2$ on each roll.

Therefore, the number of elements $= 5 \times 5 \times 5 \times 5 \times ...\left( {k - 1} \right){\text{ times}} = {5^{k - 1}}$.

Ans: Given: A die is rolled until a $2$ appears.

consider different situations where $2$ may appear in the first or second or third time etc. Add the number of elements of different situations.

In this case $2$ appears before the ${k^{th}}$ roll, so it may appear on the first roll or second roll or third role or … ${\left( {k - 1} \right)^{th}}$ roll.

If it appears on the first roll then the number of elements $= 1$.

If it appears on the second roll that means in the first roll there are $5$ outcomes and in the second role $1$, so the number of elements $= 5 \times 1$.

If it appears on the third roll that means in the first two rolls there are $5$ outcomes each and in the third role $1$, so the number of elements $= 5 \times 5 \times 1$.

Observing the pattern, the number of elements in $\left( {k - 1} \right)$ rolls $= 5 \times 5 \times 5 \times 5 \times ...\left( {k - 1} \right){\text{ times}} \times 1$.

Hence, the total number of elements of all the different situations = $1 + 5 + {5^2} + {5^3} + {...5^{\left( {k - 1} \right)}}$.

This is a geometric progression whose sum of the terms $= \left( {\dfrac{{{5^k} - 1}}{4}} \right)$.

Ans: Given: Number of times the die is rolled $= 1$

Probability of occurrence of an odd number is twice that of an even number.

$G$ is the event of occurrence of a number greater than $3$ on a single roll.

Find the total number of sample spaces when a die is rolled. Consider the odd numbers likely to occur twice.

Find the favourable sample spaces by considering the numbers $4$, $5$ or $6$ to occur.

When a die is rolled the number of outcomes is $6$ which are $\left\{ {1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6} \right\}$. It is said that each odd number is likely to occur twice of the occurrence of each even number.

So the sample spaces can be $\left\{ {1,{\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}5,{\text{ }}6} \right\}$. Here $n\left( S \right) = 9$.

The numbers greater than $3$ are $4$, $5$ and $6$, so the favourable sample spaces to occur are $\left\{ {4,{\text{ }}5,{\text{ }}5,{\text{ }}6} \right\}$. Here $n\left( E \right) = 4$.

Hence, the required probability $P\left( G \right) = \dfrac{{n\left( S \right)}}{{n\left( E \right)}} = \dfrac{4}{9}$.

Ans: Given: Probability that the chosen family owns a colour television set $= 0.87$

Probability that the chosen family owns a black and white television set $= 0.36$

Probability that the chosen family owns both the sets $= 0.30$

Assume $P\left( C \right)$ and $P\left( B \right)$ as the probabilities that the chosen family owns a colour television set and a black and white television set respectively.

Use the formula $P\left( {B \cup C} \right) = P\left( B \right) + P\left( C \right) - P\left( {B \cap C} \right)$.

Probability that the chosen family owns a colour television set $= P\left( C \right) = 0.87$.

Probability that the chosen family owns a black and white television set $= P\left( B \right) = 0.36$.

Probability that the chosen family owns both the sets $= P\left( {B \cap C} \right) = 0.30$.

Using the formula $P\left( {B \cup C} \right) = P\left( B \right) + P\left( C \right) - P\left( {B \cap C} \right)$, the probability that the chosen family owns either anyone or both kinds of sets

$= P\left( {B \cup C} \right)$

$= P\left( B \right) + P\left( C \right) - P\left( {B \cap C} \right)$

$= 0.87 + 0.36 - 0.30$

$= 0.93$ 

Ans: Given: Two mutually exclusive events $A$ and $B$.

$P\left( A \right) = 0.35$

$P\left( B \right) = 0.45$

Use the formulas: -

$P\left( {A'} \right) = 1 - P\left( A \right)$

$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$

$P\left( {A \cap B} \right) = 0$ (for mutually exclusive events)

$P\left( {A \cap B'} \right) = P\left( A \right) - P\left( {A \cap B} \right)$

$P\left( {A' \cap B'} \right) = P\left( {A \cup B} \right)'$

Here $P\left( {A'} \right)$ means the probability that the event $A$ does not occur.

 $\Rightarrow P\left( {A'} \right) = 1 - P\left( A \right)$

 $\Rightarrow P\left( {A'} \right) = 1 - 0.35$

 $\Rightarrow P\left( {A'} \right) = 0.65$ 

$\left( b \right)P\left( {B'} \right)$

Ans: Given: Two mutually exclusive events $A$ and $B$.

$P\left( A \right) = 0.35$

$P\left( B \right) = 0.45$

Use the formulas: -

$P\left( {A'} \right) = 1 - P\left( A \right)$

$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$

$P\left( {A \cap B} \right) = 0$ (for mutually exclusive events)

$P\left( {A \cap B'} \right) = P\left( A \right) - P\left( {A \cap B} \right)$

$P\left( {A' \cap B'} \right) = P\left( {A \cup B} \right)'$

Here $P\left( {B'} \right)$ means the probability that the event $B$ does not occur.

 $\Rightarrow P\left( {B'} \right) = 1 - P\left( B \right)$

 $\Rightarrow P\left( {B'} \right) = 1 - 0.45$

 $\Rightarrow P\left( {B'} \right) = 0.55$ 

$\left( c \right)P\left( {A\cup B} \right)$

Ans: Given: Two mutually exclusive events $A$ and $B$.

$P\left( A \right) = 0.35$

$P\left( B \right) = 0.45$

Use the formulas: -

$P\left( {A'} \right) = 1 - P\left( A \right)$

$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$

$P\left( {A \cap B} \right) = 0$ (for mutually exclusive events)

$P\left( {A \cap B'} \right) = P\left( A \right) - P\left( {A \cap B} \right)$

$P\left( {A' \cap B'} \right) = P\left( {A \cup B} \right)'$

Here $P\left( {A \cup B} \right)$ is the probability of occurrence of either $A$ or $B$. Using the formula $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$, where $P\left( {A \cap B} \right)$ is the probability of occurrence of $A$ and $B$ simultaneously.

$\Rightarrow P\left( {A \cup B} \right) = 0.45 + 0.35 - P\left( {A \cap B} \right)$

$\Rightarrow P\left( {A \cup B} \right) = 0.80 - P\left( {A \cap B} \right)$ 

Since the events $A$ and $B$ are mutually exclusive that means they cannot occur simultaneously. So $P\left( {A \cap B} \right) = 0$.

$\Rightarrow P\left( {A \cup B} \right) = 0.80$

$\left( d \right)P\left( {A \cap B} \right)$

Ans: Given: Two mutually exclusive events $A$ and $B$.

$P\left( A \right) = 0.35$

$P\left( B \right) = 0.45$

Use the formulas: -

$P\left( {A'} \right) = 1 - P\left( A \right)$

$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$

$P\left( {A \cap B} \right) = 0$ (for mutually exclusive events)

$P\left( {A \cap B'} \right) = P\left( A \right) - P\left( {A \cap B} \right)$

$P\left( {A' \cap B'} \right) = P\left( {A \cup B} \right)'$

Since the events $A$ and $B$ are mutually exclusive that means they cannot occur simultaneously. So $P\left( {A \cap B} \right) = 0$.

$\left( e \right)P\left( {A \cap B'} \right)$

Ans: Given: Two mutually exclusive events $A$ and $B$.

$P\left( A \right) = 0.35$

$P\left( B \right) = 0.45$

Use the formulas: -

$P\left( {A'} \right) = 1 - P\left( A \right)$

$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$

$P\left( {A \cap B} \right) = 0$ (for mutually exclusive events)

$P\left( {A \cap B'} \right) = P\left( A \right) - P\left( {A \cap B} \right)$

$P\left( {A' \cap B'} \right) = P\left( {A \cup B} \right)'$

Using the formula $P\left( {A \cap B'} \right) = P\left( A \right) - P\left( {A \cap B} \right)$,

$\Rightarrow P\left( {A \cap B'} \right) = 0.35 - 0$

$\Rightarrow P\left( {A \cap B'} \right) = 0.35$ 

$\left( f \right)P\left( {A' \cap B'} \right)$

Ans: Given: Two mutually exclusive events $A$ and $B$.

$P\left( A \right) = 0.35$

$P\left( B \right) = 0.45$

Use the formulas: -

$P\left( {A'} \right) = 1 - P\left( A \right)$

$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$

$P\left( {A \cap B} \right) = 0$ (for mutually exclusive events)

$P\left( {A \cap B'} \right) = P\left( A \right) - P\left( {A \cap B} \right)$

$P\left( {A' \cap B'} \right) = P\left( {A \cup B} \right)'$

Using the formula $P\left( {A' \cap B'} \right) = P\left( {A \cup B} \right)'$,

 $\Rightarrow P\left( {A' \cap B'} \right) = 1 - P\left( {A \cup B} \right)$

 $\Rightarrow P\left( {A' \cap B'} \right) = 1 - 0.80$

 $\Rightarrow P\left( {A' \cap B'} \right) = 0.20$ 

Ans: Given: Probabilities that the surgery is very complex $= 0.15$

Probabilities that the surgery is complex $= 0.20$

Probabilities that the surgery is routine $= 0.31$

Probabilities that the surgery is simple $= 0.26$

Probabilities that the surgery is very simple $= 0.08$

Assume $P\left( A \right)$, $P\left( B \right)$, $P\left( C \right)$, $P\left( D \right)$ and $P\left( E \right)$ as the probabilities of different surgeries. Use the formulas of set theory to find the probabilities of different combinations.

Assuming $P\left( A \right)$, $P\left( B \right)$, $P\left( C \right)$, $P\left( D \right)$ and $P\left( E \right)$ as the probabilities of surgeries rated very complex, complex, routine, simple and very simple respectively. All the events are mutually exclusive that means no two of them can occur simultaneously.

Probability that a particular surgery is rated complex or very complex is $P\left( {A \cup B} \right)$.

 $\Rightarrow P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$

 $\Rightarrow P\left( {A \cup B} \right) = 0.15 + 0.20 - 0$

 $\Rightarrow P\left( {A \cup B} \right) = 0.35$ 

Ans: Given: Probabilities that the surgery is very complex $= 0.15$

Probabilities that the surgery is complex $= 0.20$

Probabilities that the surgery is routine $= 0.31$

Probabilities that the surgery is simple $= 0.26$

Probabilities that the surgery is very simple $= 0.08$

Assume $P\left( A \right)$, $P\left( B \right)$, $P\left( C \right)$, $P\left( D \right)$ and $P\left( E \right)$ as the probabilities of different surgeries. Use the formulas of set theory to find the probabilities of different combinations.

Assuming $P\left( A \right)$, $P\left( B \right)$, $P\left( C \right)$, $P\left( D \right)$ and $P\left( E \right)$ as the probabilities of surgeries rated very complex, complex, routine, simple and very simple respectively. All the events are mutually exclusive that means no two of them can occur simultaneously.

Probability that a particular surgery is rated neither very complex nor very simple is $P\left( {A' \cap E'} \right)$.

$\Rightarrow P\left( {A' \cap E'} \right) = 1 - P\left( {A \cup E} \right)$

$\Rightarrow P\left( {A' \cap E'} \right) = 1 - \left( {P\left( A \right) + P\left( E \right) - P\left( {A \cap E} \right)} \right)$

$\Rightarrow P\left( {A' \cap E'} \right) = 1 - \left( {0.15 + 0.08 - 0} \right)$

$\Rightarrow P\left( {A' \cap E'} \right) = 0.77$ 

Ans: Given: Probabilities that the surgery is very complex $= 0.15$

Probabilities that the surgery is complex $= 0.20$

Probabilities that the surgery is routine $= 0.31$

Probabilities that the surgery is simple $= 0.26$

Probabilities that the surgery is very simple $= 0.08$

Assume $P\left( A \right)$, $P\left( B \right)$, $P\left( C \right)$, $P\left( D \right)$ and $P\left( E \right)$ as the probabilities of different surgeries. Use the formulas of set theory to find the probabilities of different combinations.

Assuming $P\left( A \right)$, $P\left( B \right)$, $P\left( C \right)$, $P\left( D \right)$ and $P\left( E \right)$ as the probabilities of surgeries rated very complex, complex, routine, simple and very simple respectively. All the events are mutually exclusive that means no two of them can occur simultaneously.

Probability that a particular surgery is rated routine or complex is $P\left( {B \cup C} \right)$.

$\Rightarrow P\left( {B \cup C} \right) = P\left( B \right) + P\left( C \right) - P\left( {B \cap C} \right)$

$\Rightarrow P\left( {B \cup C} \right) = 0.20 + 0.31 - 0$

 $\Rightarrow P\left( {B \cup C} \right) = 0.51$

Ans: Given: Probabilities that the surgery is very complex $= 0.15$

Probabilities that the surgery is complex $= 0.20$

Probabilities that the surgery is routine $= 0.31$

Probabilities that the surgery is simple $= 0.26$

Probabilities that the surgery is very simple $= 0.08$

Assume $P\left( A \right)$, $P\left( B \right)$, $P\left( C \right)$, $P\left( D \right)$ and $P\left( E \right)$ as the probabilities of different surgeries. Use the formulas of set theory to find the probabilities of different combinations.

Assuming $P\left( A \right)$, $P\left( B \right)$, $P\left( C \right)$, $P\left( D \right)$ and $P\left( E \right)$ as the probabilities of surgeries rated very complex, complex, routine, simple and very simple respectively. All the events are mutually exclusive that means no two of them can occur simultaneously.

Probability that a particular surgery is rated routine or simple is $P\left( {C \cup D} \right)$.

$\Rightarrow P\left( {C \cup D} \right) = P\left( C \right) + P\left( D \right) - P\left( {C \cap D} \right)$

$\Rightarrow P\left( {C \cup D} \right) = 0.31 + 0.26 - 0$

$\Rightarrow P\left( {C \cup D} \right) = 0.57$ 

Ans: Given: Total number of candidates $= 4\left\{ {A,{\text{ }}B,{\text{ }}C,{\text{ }}D} \right\}$

Probability of selection of $A =$$2 \times$ Probability of selection of $B$

Probability of selection of $B =$ Probability of selection of $C$

Probability of selection of $C =$$2 \times$ Probability of selection of $D$

Assume $P\left( A \right)$, $P\left( B \right)$, $P\left( C \right)$ and $P\left( D \right)$ as the probabilities of selection of the given candidates. Form relations between them and use the formula $P\left( A \right) + P\left( B \right) + P\left( C \right) + P\left( D \right) = 1$ to find their values.

Assuming $P\left( A \right)$, $P\left( B \right)$, $P\left( C \right)$ and $P\left( D \right)$ as the probabilities of selection of the candidates $A$, $B$, $C$ and $D$ respectively. All the events are mutually exclusive that means no two of them can occur simultaneously.

According to the information given,

$P\left( A \right) = 2P\left( B \right)$ … (1)

$P\left( B \right) = P\left( C \right)$ … (2)

$P\left( C \right) = 2P\left( D \right)$ … (3)

Using the formula $P\left( A \right) + P\left( B \right) + P\left( C \right) + P\left( D \right) = 1$ and solving the three equations,

$P\left( A \right) = \dfrac{4}{9}$, $P\left( B \right) = \dfrac{2}{9}$, $P\left( C \right) = \dfrac{2}{9}$ and $P\left( D \right) = \dfrac{1}{9}$

Probability that $C$ will be selected is $P\left( C \right) = \dfrac{2}{9}$.

Ans: Given: Total number of candidates $= 4\left\{ {A,{\text{ }}B,{\text{ }}C,{\text{ }}D} \right\}$

Probability of selection of $A =$ $2 \times$ Probability of selection of $B$

Probability of selection of $B =$ Probability of selection of $C$

Probability of selection of $C =$ $2 \times$ Probability of selection of $D$

Assume $P\left( A \right)$, $P\left( B \right)$, $P\left( C \right)$ and $P\left( D \right)$ as the probabilities of selection of the given candidates. Form relations between them and use the formula $P\left( A \right) + P\left( B \right) + P\left( C \right) + P\left( D \right) = 1$ to find their values.

Assuming $P\left( A \right)$, $P\left( B \right)$, $P\left( C \right)$ and $P\left( D \right)$ as the probabilities of selection of the candidates $A$, $B$, $C$ and $D$ respectively. All the events are mutually exclusive that means no two of them can occur simultaneously.

According to the information given,

$P\left( A \right) = 2P\left( B \right)$ … (1)

$P\left( B \right) = P\left( C \right)$ … (2)

$P\left( C \right) = 2P\left( D \right)$ … (3)

Using the formula $P\left( A \right) + P\left( B \right) + P\left( C \right) + P\left( D \right) = 1$ and solving the three equations,

$P\left( A \right) = \dfrac{4}{9}$, $P\left( B \right) = \dfrac{2}{9}$, $P\left( C \right) = \dfrac{2}{9}$ and $P\left( D \right) = \dfrac{1}{9}$

Probability that $A$ will not be selected is $P\left( {A'} \right) = 1 - P\left( A \right)$.

 $\Rightarrow P\left( {A'} \right) = 1 - \dfrac{4}{9}$

 $\Rightarrow P\left( {A'} \right) = \dfrac{5}{9}$ 

Ans:Given:Total number of persons $= 4$

Probability of John’s promotion $=$ Probability of Gurpreet’s promotion

Probability of Rita’s promotion $=$$2 \times$ Probability of John’s promotion

Probability of Aslam’s promotion $=$$4 \times$ Probability of John’s promotion

Assume $P\left( W \right)$, $P\left( X \right)$, $P\left( Y \right)$ and $P\left( Z \right)$ as the probabilities of promotion of the four persons. Form relations between them and use the formula $P\left( W \right) + P\left( X \right) + P\left( Y \right) + P\left( Z \right) = 1$ to find their values.

Assuming $P\left( W \right)$, $P\left( X \right)$, $P\left( Y \right)$ and $P\left( Z \right)$  as the probabilities of promotion of John, Rita, Aslam and Gurpreet respectively. All the events are mutually exclusive that means no two of them can occur simultaneously.

According to the information given,

$P\left( W \right) = P\left( Z \right)$ … (1)

$P\left( X \right) = 2P\left( W \right)$ … (2)

$P\left( Y \right) = 4P\left( W \right)$ … (3)

Using the formula $P\left( W \right) + P\left( X \right) + P\left( Y \right) + P\left( Z \right) = 1$ and solving the three equations,

$P\left( W \right) = \dfrac{1}{8}$, $P\left( X \right) = \dfrac{1}{4}$, $P\left( Y \right) = \dfrac{1}{2}$ and $P\left( Z \right) = \dfrac{1}{8}$

Probability of promotion of John is $P\left( W \right) = \dfrac{1}{8}$.

Probability of promotion of Rita is $P\left( X \right) = \dfrac{1}{4}$.

Probability of promotion of Aslam is $P\left( Y \right) = \dfrac{1}{2}$.

Probability of promotion of Gurpreet is $P\left( Z \right) = \dfrac{1}{8}$.

Ans: Given: Total number of persons $= 4$

Probability of John’s promotion $=$ Probability of Gurpreet’s promotion

Probability of Rita’s promotion $=$$2 \times$ Probability of John’s promotion

Probability of Aslam’s promotion $=$$4 \times$ Probability of John’s promotion

Assume $P\left( W \right)$, $P\left( X \right)$, $P\left( Y \right)$ and $P\left( Z \right)$ as the probabilities of promotion of the four persons. Form relations between them and use the formula $P\left( W \right) + P\left( X \right) + P\left( Y \right) + P\left( Z \right) = 1$ to find their values.

Assuming $P\left( W \right)$, $P\left( X \right)$, $P\left( Y \right)$ and $P\left( Z \right)$  as the probabilities of promotion of John, Rita, Aslam and Gurpreet respectively. All the events are mutually exclusive that means no two of them can occur simultaneously.

According to the information given,

$P\left( W \right) = P\left( Z \right)$ … (1)

$P\left( X \right) = 2P\left( W \right)$ … (2)

$P\left( Y \right) = 4P\left( W \right)$ … (3)

Using the formula $P\left( W \right) + P\left( X \right) + P\left( Y \right) + P\left( Z \right) = 1$ and solving the three equations,

$P\left( W \right) = \dfrac{1}{8}$, $P\left( X \right) = \dfrac{1}{4}$, $P\left( Y \right) = \dfrac{1}{2}$ and $P\left( Z \right) = \dfrac{1}{8}$

Probability of promotion of John or Gurpreet $P\left( A \right) = P\left( {W \cup Z} \right)$.

$\Rightarrow P\left( A \right) = P\left( W \right) + P\left( Z \right) - P\left( {W \cap Z} \right)$

 $\Rightarrow P\left( A \right) = \dfrac{1}{8} + \dfrac{1}{8} - 0$

 $\Rightarrow P\left( A \right) = \dfrac{1}{4}$