Chapter 16: Probability

Ans: Given:

In the Venn diagram: -

$P\left( {{\text{only }}A} \right) = 0.13$

$P\left( {{\text{only }}B} \right) = 0.10$

\[P\left( {{\text{only }}C} \right) = 0.28\]

$P\left( {A \cap B} \right) = 0.07$

$P\left( {B \cap C} \right) = 0.15$

Using the Venn and the formulas of theory solve for each subpart of the question.

Use the formulas: -

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

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

The probability of occurrence of the event $A$ is \[P\left( A \right) = P\left( {{\text{only }}A} \right) + P\left( {A \cap B} \right)\].

 $\Rightarrow P\left( A \right) = 0.13 + 0.07$

 $\Rightarrow P\left( A \right) = 0.20$ 

$\left( b \right)P\left( {B \cap \overline C } \right)$

Ans: Given:

In the Venn diagram: -

$P\left( {{\text{only }}A} \right) = 0.13$

$P\left( {{\text{only }}B} \right) = 0.10$

\[P\left( {{\text{only }}C} \right) = 0.28\]

$P\left( {A \cap B} \right) = 0.07$

$P\left( {B \cap C} \right) = 0.15$

Using the Venn and the formulas of theory solve for each subpart of the question.

Use the formulas: -

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

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

Here $\left( {B \cap \overline C } \right)$ states the event of occurrence of $B$ and not $C$.

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

 $\Rightarrow P\left( {B \cap \overline C } \right) = \left( {0.10 + 0.07 + 0.15} \right) - 0.15$

 $\Rightarrow P\left( {B \cap \overline C } \right) = 0.17$ 

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

Ans: Given:

In the Venn diagram: -

$P\left( {{\text{only }}A} \right) = 0.13$

$P\left( {{\text{only }}B} \right) = 0.10$

\[P\left( {{\text{only }}C} \right) = 0.28\]

$P\left( {A \cap B} \right) = 0.07$

$P\left( {B \cap C} \right) = 0.15$

Using the Venn and the formulas of theory solve for each subpart of the question.

Use the formulas: -

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

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

Here $P\left( {A \cup B} \right)$ is the probability of occurrence of either $A$ or $B$.

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

 $\Rightarrow P\left( {A \cup B} \right) = \left( {0.13 + 0.07} \right) + \left( {0.10 + 0.07 + 0.15} \right) - 0.07$

 $\Rightarrow P\left( {A \cup B} \right) = 0.45$ 

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

Ans: Given:

In the Venn diagram: -

$P\left( {{\text{only }}A} \right) = 0.13$

$P\left( {{\text{only }}B} \right) = 0.10$

\[P\left( {{\text{only }}C} \right) = 0.28\]

$P\left( {A \cap B} \right) = 0.07$

$P\left( {B \cap C} \right) = 0.15$

Using the Venn and the formulas of theory solve for each subpart of the question.

Use the formulas: -

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

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

Here $\left( {A \cap \overline B } \right)$ states the event of occurrence of $A$ and not $B$.

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

 $\Rightarrow P\left( {A \cap \overline B } \right) = \left( {0.13 + 0.07} \right) - 0.07$

 $\Rightarrow P\left( {A \cap \overline B } \right) = 0.13$ 

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

Ans: Given:

In the Venn diagram: -

$P\left( {{\text{only }}A} \right) = 0.13$

$P\left( {{\text{only }}B} \right) = 0.10$

\[P\left( {{\text{only }}C} \right) = 0.28\]

$P\left( {A \cap B} \right) = 0.07$

$P\left( {B \cap C} \right) = 0.15$

Using the Venn and the formulas of theory solve for each subpart of the question.

Use the formulas: -

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

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

Here $\left( {B \cap C} \right)$ states the event of occurrence of both $A$ and $B$.

$\Rightarrow P\left( {B \cap C} \right) = 0.15$

$\left( f \right)$ Probability of exactly one of the three occurs.

Ans: Given:

In the Venn diagram: -

$P\left( {{\text{only }}A} \right) = 0.13$

$P\left( {{\text{only }}B} \right) = 0.10$

\[P\left( {{\text{only }}C} \right) = 0.28\]

$P\left( {A \cap B} \right) = 0.07$

$P\left( {B \cap C} \right) = 0.15$

Using the Venn and the formulas of theory solve for each subpart of the question.

Use the formulas: -

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

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

Probability of exactly one of the three occurs means \[P\left( {{\text{only }}A} \right) + P\left( {{\text{only }}B} \right) + P\left( {{\text{only }}C} \right)\].

$\Rightarrow $ Required probability $= 0.13 + 0.10 + 0.28$

$\Rightarrow $ Required probability $= 0.51$

Ans: Given: Two black balls (labelled $B1$ and $B2$) and one white ball in the first urn.

Two white balls (labelled $W1$ and $W2$) and one black ball in the second urn.

One of the two urns is chosen then a ball is chosen and then a second ball is chosen without replacement.

Find all the possible combinations that can be made in selecting the two balls one after the other and not at once. Check the favourable number of sample spaces for the subparts.

Let us represent the single white ball in the first urn as $W$ and the single black ball in the second urn as $B$. The possible combinations of selecting the two balls one after the other from the urns is the total number of sample spaces given as,

\[n\left( S \right) = \left\{ {B1B2,{\text{ }}B2B1,{\text{ }}B1W,{\text{ }}WB1,{\text{ }}B2W,{\text{ }}WB2,{\text{ }}W1W2,{\text{ }}W2W1,{\text{ }}BW1,{\text{ }}W1B,{\text{ }}BW2,{\text{ }}W2B} \right\} = 12\]

(b) What is the probability that two black balls are chosen?

Ans: Given: Two black balls (labelled $B1$ and $B2$) and one white ball in the first urn.

Two white balls (labelled $W1$ and $W2$) and one black ball in the second urn.

One of the two urns is chosen then a ball is chosen and then a second ball is chosen without replacement.

Find all the possible combinations that can be made in selecting the two balls one after the other and not at once. Check the favourable number of sample spaces for the subparts.

The combinations where two blacks balls are chosen is $n\left( X \right) = \left\{ {B1B2,{\text{ }}B2B1} \right\} = 2$.

Therefore, the required probability is $\dfrac{{n\left( X \right)}}{{n\left( S \right)}} = \dfrac{2}{{12}} = \dfrac{1}{6}$.

Ans: Given: Two black balls (labelled $B1$ and $B2$) and one white ball in the first urn.

Two white balls (labelled $W1$ and $W2$) and one black ball in the second urn.

One of the two urns is chosen then a ball is chosen and then a second ball is chosen without replacement.

Find all the possible combinations that can be made in selecting the two balls one after the other and not at once. Check the favourable number of sample spaces for the subparts.

The combinations where two balls of opposite colours are chosen is $n\left( Y \right) = \left\{ {B1W,{\text{ }}WB1,{\text{ }}B2W,{\text{ }}WB2,BW1,{\text{ }}W1B,{\text{ }}BW2,{\text{ }}W2B} \right\} = 8$.

Therefore, the required probability is $\dfrac{{n\left( Y \right)}}{{n\left( S \right)}} = \dfrac{8}{{12}} = \dfrac{2}{3}$.

Ans: Given: Total number of red balls $= 8$

Total number of white balls $= 5$

Three balls are drawn at random.

Use the formula ${}^{13}{C_3}$ to find the total number of sample spaces.

Select the balls using the formula of combination for each subpart.

Since there are a total of $13$ balls and any $3$ balls can be selected in ${}^{13}{C_3}$ ways, so the total number of sample space is $n\left( S \right) = {}^{13}{C_3}$.

To select all the three white balls they must be selected from the given $5$ white balls, so the number of favourable sample spaces is $n\left( A \right) = {}^5{C_3}$.

Therefore, the required probability is $= \dfrac{{n\left( A \right)}}{{n\left( S \right)}} = \dfrac{{{}^5{C_3}}}{{{}^{13}{C_3}}} = \dfrac{5}{{143}}$.

(b) All the three balls are red.

Ans: Given: Total number of red balls $= 8$

Total number of white balls $= 5$

Three balls are drawn at random.

Use the formula ${}^{13}{C_3}$ to find the total number of sample spaces.

Select the balls using the formula of combination for each subpart.

Since there are a total of $13$ balls and any $3$ balls can be selected in ${}^{13}{C_3}$ ways, so the total number of sample space is $n\left( S \right) = {}^{13}{C_3}$.

To select all the three red balls they must be selected from the given $8$ red balls, so the number of favourable sample spaces is $n\left( B \right) = {}^8{C_3}$.

Therefore, the required probability is $= \dfrac{{n\left( B \right)}}{{n\left( S \right)}} = \dfrac{{{}^8{C_3}}}{{{}^{13}{C_3}}} = \dfrac{{28}}{{143}}$.

Ans: Given: Total number of red balls $= 8$

Total number of white balls $= 5$

Three balls are drawn at random.

Use the formula ${}^{13}{C_3}$ to find the total number of sample spaces.

Select the balls using the formula of combination for each subpart.

Since there are a total of $13$ balls and any $3$ balls can be selected in ${}^{13}{C_3}$ ways, so the total number of sample space is $n\left( S \right) = {}^{13}{C_3}$.

To select one red ball the number of ways is \[{}^8{C_1}\] and to select two white balls the number of ways is \[{}^5{C_2}\]. So the number of favourable sample spaces is $n\left( C \right) = {}^8{C_1} \times {}^5{C_2}$.

Therefore, the required probability is $= \dfrac{{n\left( C \right)}}{{n\left( S \right)}} = \dfrac{{{}^8{C_1} \times {}^5{C_2}}}{{{}^{13}{C_3}}} = \dfrac{{40}}{{143}}$.

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

Find the total number of sample spaces by arranging all the thirteen letters of the word and dividing it by the product of the factorial of the number of times each letter is repeating.

Arrange the letters as per the condition given in the subparts to find the favourable sample spaces.

There are $13$ letters in the word ASSASSINATION in which A is repeated $3$ times, S is repeated $4$ times, I is repeated $2$ times and N is repeated $2$ times. So the number of different words that can be formed is the total number of sample spaces $n\left( S \right) = \dfrac{{13!}}{{4!{\text{ }} \times {\text{ }}3!{\text{ }} \times {\text{ }}2!{\text{ }} \times {\text{ }}2!}}$.

All the four S’s come together so they can be assumed as a single unit. Since, they are identical alphabets, so there is no meaning of arranging them among themselves. The remaining $9$ alphabets and this single unit can be considered as $10$ objects and they can be arranged in $n\left( A \right) = \dfrac{{10!}}{{3!{\text{ }} \times {\text{ }}2!{\text{ }} \times {\text{ }}2!}}$ ways.

Therefore, the required probability is $\dfrac{{n\left( A \right)}}{{n\left( S \right)}} = \dfrac{2}{{143}}$.

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

Find the total number of sample spaces by arranging all the thirteen letters of the word and dividing it by the product of the factorial of the number of times each letter is repeating.

Arrange the letters as per the condition given in the subparts to find the favourable sample spaces.

There are $13$ letters in the word ASSASSINATION in which A is repeated $3$ times, S is repeated $4$ times, I is repeated $2$ times and N is repeated $2$ times. So the number of different words that can be formed is the total number of sample spaces $n\left( S \right) = \dfrac{{13!}}{{4!{\text{ }} \times {\text{ }}3!{\text{ }} \times {\text{ }}2!{\text{ }} \times {\text{ }}2!}}$.

Here two I’s and two N’s come together (IINN). These four letters can be arranged in $\dfrac{{4!}}{{2!{\text{ }} \times {\text{ }}2!}}$ ways among themselves. The remaining $9$ alphabets and this single unit can be considered as $10$ objects and they can be arranged in $\dfrac{{10!}}{{4! \times 3!}}$ ways. So the number of favorable arrangements is $n\left( B \right) = \dfrac{{4!}}{{2!{\text{ }} \times {\text{ }}2!}}{\text{ }} \times {\text{ }}\dfrac{{10!}}{{4!{\text{ }} \times {\text{ }}3!}}$.

Therefore, the required probability is $\dfrac{{n\left( B \right)}}{{n\left( S \right)}} = \dfrac{2}{{143}}$.

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

Find the total number of sample spaces by arranging all the thirteen letters of the word and dividing it by the product of the factorial of the number of times each letter is repeating.

Arrange the letters as per the condition given in the subparts to find the favourable sample spaces.

There are $13$ letters in the word ASSASSINATION in which A is repeated $3$ times, S is repeated $4$ times, I is repeated $2$ times and N is repeated $2$ times. So the number of different words that can be formed is the total number of sample spaces $n\left( S \right) = \dfrac{{13!}}{{4!{\text{ }} \times {\text{ }}3!{\text{ }} \times {\text{ }}2!{\text{ }} \times {\text{ }}2!}}$.

Let us assume that all the three A’s come together so they can be assumed as a single unit. They are identical alphabets. The remaining $10$ alphabets and this single unit can be considered as $11$ objects and they can be arranged in $n\left( C \right) = \dfrac{{11!}}{{4!{\text{ }} \times {\text{ }}2!{\text{ }} \times {\text{ }}2!}}$ ways.

Therefore, the probability that all the A’s comes together is $\dfrac{{n\left( C \right)}}{{n\left( S \right)}} = \dfrac{1}{{26}}$.

Hence, the probability that all the A’s does not come together is $\left( {1 - \dfrac{{n\left( C \right)}}{{n\left( S \right)}}} \right) = \dfrac{{25}}{{26}}$.

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

Find the total number of sample spaces by arranging all the thirteen letters of the word and dividing it by the product of the factorial of the number of times each letter is repeating.

Arrange the letters as per the condition given in the subparts to find the favourable sample spaces.

There are $13$ letters in the word ASSASSINATION in which A is repeated $3$ times, S is repeated $4$ times, I is repeated $2$ times and N is repeated $2$ times. So the number of different words that can be formed is the total number of sample spaces $n\left( S \right) = \dfrac{{13!}}{{4!{\text{ }} \times {\text{ }}3!{\text{ }} \times {\text{ }}2!{\text{ }} \times {\text{ }}2!}}$.

Here no two A’s come together. Let us arrange the remaining letters first.

_ S _ S _ S _ S _ I _ N _ T _ I _ O _ N _

The three A’s can be filled at any $3$ of the $11$ blank spaces in ${}^{11}{C_3}$ ways (arrangement not required as they are identical). The remaining $10$ letters can be arranged in $\dfrac{{10!}}{{4!{\text{ }} \times {\text{ }}2!{\text{ }} \times {\text{ }}2!}}$ ways. So the number of favourable arrangements is $n\left( D \right) = {}^{11}{C_3} \times \dfrac{{10!}}{{4!{\text{ }} \times {\text{ }}2!{\text{ }} \times {\text{ }}2!}}$.

Therefore, the required probability is $\dfrac{{n\left( D \right)}}{{n\left( S \right)}} = \dfrac{{15}}{{26}}$.

Ans: Given: A deck of $52$ cards and one card is drawn.

Assume $P\left( K \right)$, $P\left( H \right)$ and $P\left( R \right)$ as the probabilities of drawing a king, a heart and a red card respectively.

Use the formula $P\left( {K \cup H \cup R} \right) = P\left( K \right) + P\left( H \right) + P\left( R \right) - P\left( {K \cap H} \right) - P\left( {H \cap R} \right) - P\left( {K \cap R} \right) + P\left( {K \cap H \cap R} \right)$ to solve for the $L.H.S$.

Let us assume $P\left( K \right)$, $P\left( H \right)$ and $P\left( R \right)$ as the probabilities of drawing a king, a heart and a red card respectively. The number of ways to draw a $1$ card from $52$ cards is $n\left( S \right) = {}^{52}{C_1}$.

There are four kings, thirteen hearts cards and 26 red cards.

Now, there is $1$ card which is both king and red, $13$ cards which are both red and hearts, $2$ cards which are both red and king.

There is only one card which is king, hearts and red.

So the probability of drawing a king, a heart or a red card is $P\left( {K \cup H \cup R} \right)$ which is

$= P\left( K \right) + P\left( H \right) + P\left( R \right) - P\left( {K \cap H} \right) - P\left( {H \cap R} \right) - P\left( {K \cap R} \right) + P\left( {K \cap H \cap R} \right)$

$= \dfrac{4}{{52}} + \dfrac{{13}}{{52}} + \dfrac{{26}}{{52}} - \dfrac{1}{{52}} - \dfrac{{13}}{{52}} - \dfrac{2}{{52}} + \dfrac{1}{{52}}$

$= \dfrac{{28}}{{52}}$

$= \dfrac{7}{{13}}$ 

Ans: Given: The $9$ elementary outcomes ${e_1}$, ${e_2}$, …, ${e_9}$ and their probabilities.

$A = \left\{ {{e_1},{\text{ }}{e_2},{\text{ }}{e_8}} \right\}$

$B = \left\{ {{e_2},{\text{ }}{e_5},{\text{ }}{e_8},{\text{ }}{e_9}} \right\}$

Use the formulas: -

$P\left( A \right) = P\left( {{e_1}} \right) + P\left( {{e_2}} \right) + P\left( {{e_8}} \right)$ and $P\left( B \right) = P\left( {{e_2}} \right) + P\left( {{e_5}} \right) + P\left( {{e_8}} \right) + P\left( {{e_9}} \right)$

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

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

Since $A = \left\{ {{e_1},{\text{ }}{e_2},{\text{ }}{e_8}} \right\}$, therefore,

 $\Rightarrow P\left( A \right) = P\left( {{e_1}} \right) + P\left( {{e_2}} \right) + P\left( {{e_8}} \right)$

 $\Rightarrow P\left( A \right) = 0.08 + 0.08 + 0.07$

 $\Rightarrow P\left( A \right) = 0.23$ 

Since $B = \left\{ {{e_2},{\text{ }}{e_5},{\text{ }}{e_8},{\text{ }}{e_9}} \right\}$, therefore,

 $\Rightarrow P\left( B \right) = P\left( {{e_2}} \right) + P\left( {{e_5}} \right) + P\left( {{e_8}} \right) + P\left( {{e_9}} \right)$

 $\Rightarrow P\left( B \right) = 0.08 + 0.1 + 0.07 + 0.07$

 $\Rightarrow P\left( B \right) = 0.32$ 

Now, $\left( {A \cap B} \right)$ is the set of common elementary outcomes of $A$ and $B$. So $\left( {A \cap B} \right) = \left\{ {{e_5},{\text{ }}{e_8}} \right\}$, therefore,

 $\Rightarrow P\left( {A \cap B} \right) = P\left( {{e_5}} \right) + P\left( {{e_8}} \right)$

 $\Rightarrow P\left( {A \cap B} \right) = 0.1 + 0.07$

 $\Rightarrow P\left( {A \cap B} \right) = 0.17$ 

Ans: Given: The $9$ elementary outcomes ${e_1}$, ${e_2}$, …, ${e_9}$ and their probabilities.

$A = \left\{ {{e_1},{\text{ }}{e_2},{\text{ }}{e_8}} \right\}$

$B = \left\{ {{e_2},{\text{ }}{e_5},{\text{ }}{e_8},{\text{ }}{e_9}} \right\}$

Use the formulas: -

$P\left( A \right) = P\left( {{e_1}} \right) + P\left( {{e_2}} \right) + P\left( {{e_8}} \right)$ and $P\left( B \right) = P\left( {{e_2}} \right) + P\left( {{e_5}} \right) + P\left( {{e_8}} \right) + P\left( {{e_9}} \right)$

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

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

According to the addition law of probability $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$,

$\Rightarrow P\left( {A \cup B} \right) = 0.23 + 0.25 - 0.17$

$\Rightarrow P\left( {A \cup B} \right) = 0.23 + 0.32 - 0.40$ 

Ans: Given: The $9$ elementary outcomes ${e_1}$, ${e_2}$, …, ${e_9}$ and their probabilities.

$A = \left\{ {{e_1},{\text{ }}{e_2},{\text{ }}{e_8}} \right\}$

$B = \left\{ {{e_2},{\text{ }}{e_5},{\text{ }}{e_8},{\text{ }}{e_9}} \right\}$

Use the formulas: -

$P\left( A \right) = P\left( {{e_1}} \right) + P\left( {{e_2}} \right) + P\left( {{e_8}} \right)$ and $P\left( B \right) = P\left( {{e_2}} \right) + P\left( {{e_5}} \right) + P\left( {{e_8}} \right) + P\left( {{e_9}} \right)$

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

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

The event $\left( {A \cup B} \right)$ is the set of all the elements present in either $A$ or $B$. So the composition of $\left( {A \cup B} \right)$ is $\left\{ {{e_1},{\text{ }}{e_2},{\text{ }}{e_5},{\text{ }}{e_8},{\text{ }}{e_9}} \right\}$, therefore,

$\Rightarrow P\left( {A \cup B} \right) = P\left( {{e_1}} \right) + P\left( {{e_2}} \right) + P\left( {{e_5}} \right) + P\left( {{e_8}} \right) + P\left( {{e_9}} \right)$

 $\Rightarrow P\left( {A \cup B} \right) = 0.08 + 0.08 + 0.1 + 0.07 + 0.07$

 $\Rightarrow P\left( {A \cup B} \right) = 0.40$

Ans: Given: The $9$ elementary outcomes ${e_1}$, ${e_2}$, …, ${e_9}$ and their probabilities.

$A = \left\{ {{e_1},{\text{ }}{e_2},{\text{ }}{e_8}} \right\}$

$B = \left\{ {{e_2},{\text{ }}{e_5},{\text{ }}{e_8},{\text{ }}{e_9}} \right\}$

Use the formulas: -

$P\left( A \right) = P\left( {{e_1}} \right) + P\left( {{e_2}} \right) + P\left( {{e_8}} \right)$ and $P\left( B \right) = P\left( {{e_2}} \right) + P\left( {{e_5}} \right) + P\left( {{e_8}} \right) + P\left( {{e_9}} \right)$

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

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

Using $P\left( B \right)$ the value of $P\left( {\overline B } \right)$ is given as $1 - P\left( B \right)$.

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

 $\Rightarrow P\left( {\overline B } \right) = 1 - 0.32$

 $\Rightarrow P\left( {\overline B } \right) = 0.68$ 

Since $B = \left\{ {{e_2},{\text{ }}{e_5},{\text{ }}{e_8},{\text{ }}{e_9}} \right\}$, so $\overline B  = \left\{ {{e_1},{\text{ }}{e_3},{\text{ }}{e_4},{\text{ }}{e_6},{\text{ }}{e_7}} \right\}$, therefore,

 $\Rightarrow P\left( {\overline B } \right) = P\left( {{e_1}} \right) + P\left( {{e_3}} \right) + P\left( {{e_4}} \right) + P\left( {{e_6}} \right) + P\left( {{e_7}} \right)$

 $\Rightarrow P\left( {\overline B } \right) = 0.08 + 0.1 + 0.1 + 0.2 + 0.2$

 $\Rightarrow P\left( {\overline B } \right) = 0.68$ 

Ans: write all the sample space and find the favourable events among them.

When the fair die is tossed one time then the sample spaces are $n\left( S \right) = \left\{ {1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6} \right\} = 6$. There are three odd numbers, so the favourable events are $n\left( A \right) = \left\{ {1,{\text{ }}3,{\text{ }}5} \right\} = 3$.

Therefore, the probability $p$ is $\dfrac{{n\left( A \right)}}{{n\left( S \right)}} = \dfrac{3}{6} = \dfrac{1}{2}$.

Ans: write all the sample space and find the favourable events among them.

When two fair coins are tossed the sample spaces are $n\left( S \right) = \left\{ {HH,{\text{ }}HT,{\text{ }}TH,{\text{ }}TT} \right\} = 4$.

There are three outcomes where at least one head appears, so $n\left( B \right) = \left\{ {HH,{\text{ }}HT,{\text{ }}TH} \right\} = 3$.

Therefore, the probability $p$ is $\dfrac{{n\left( B \right)}}{{n\left( S \right)}} = \dfrac{3}{4}$.

Ans: consider that there are four kings, one $9$ of hearts and one $3$ of spades in a pack.

In a deck of $52$ cards the number of ways to draw a card is the total number of sample spaces given as $n\left( S \right) = {}^{52}{C_1} = 52$.

There are four kings (one in each suit), one $9$ of hearts and one $3$ of spades in a pack. So the number of favourable outcomes is $n\left( C \right) = \left( {4 + 1 + 1} \right) = 6$.

Therefore, the probability $p$ is $\dfrac{{n\left( C \right)}}{{n\left( S \right)}} = \dfrac{6}{{52}} = \dfrac{3}{{26}}$.

Ans: write all the sample space and find the favourable events among them.

When a pair of dice is tossed one time then there are ${6^2} = 36$ combinations of outcomes possible. So the number of sample spaces is $n\left( S \right) = 36$.

The sum of $6$ appears in the cases $\left( {1,5} \right)$, $\left( {5,1} \right)$, $\left( {2,4} \right)$, $\left( {4,2} \right)$ and $\left( {3,3} \right)$, so the number of favourable outcomes is $n\left( D \right) = 5$.

Therefore, the probability $p$ is $\dfrac{{n\left( D \right)}}{{n\left( S \right)}} = \dfrac{5}{{36}}$.

Ans: Given: Non-leap year, There are  \[365\] days that have \[52\] weeks and \[1\] days.

$P\left( A \right){\text{ }} = \dfrac{{\left\{ {{\text{Tuesday}}} \right\}}}{{\left\{ {S{\text{unday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}}} \right\}}}$

$P\left( B \right){\text{ }} = \dfrac{{\left\{ {{\text{Wednesday}}} \right\}}}{{\left\{ {{\text{Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}}} \right\}}}$ 

If this day is a Tuesday or Wednesday then, the year will have\[53\]Tuesday

Or \[53\] Wednesday.

\[\therefore \] Required Probability \[ = \dfrac{1}{7} + \dfrac{1}{7} = \dfrac{2}{7}\]

Correct Answer: B

Ans: Given: Three numbers are chosen from  \[1{\text{ }}\] to \[20\]

The probability that the numbers are not a consecutive \[ = 1- \] probability of the numbers being consecutive.

Since the set of consecutive numbers are from$1{  -  }20$

  ${\text{P}}\left( {1,{\text{ }}2,{\text{ }}3} \right),{\text{ }}\left( {2,{\text{ }}3{\text{ }},4} \right){\text{ }},\left( {3,{\text{ }}4,{\text{ }}5} \right){\text{ }}........({\text{ }}18,{\text{ }}19,{\text{ }}20){\text{i}}{\text{.e}}.,{\text{ }}18.$

  ${\text{P }}\left( {{\text{ consecutive numbers }}} \right){\text{  =  }}\dfrac{{18}}{{^{20}{C_3}}}$

   $= {\text{ }}\dfrac{{18}}{{\dfrac{{20{ } \times{ }19{ } \times{ }18}}{{3{ }!}}}}{\text{ }} = {\text{ }}\dfrac{3}{{19}}$

  ${\text{P }}\left( {{\text{ three number are not consecutive}}} \right){\text{ }} = {\text{ }}1{\text{ }} - {\text{ }}\dfrac{3}{{190}}{\text{ }} = {\text{ }}\dfrac{{187}}{{190}}$ 

Correct Answer: B

Ans: Given: While shuffling a pack, \[52\] Playing cards and \[2\]  got accidentally dropped.

\[{\text{P}}\left( {{\text{both cards of opposite colour}}} \right) = \dfrac{{^{26}{C_1}\times {{ }^{26}}{C_1}}}{{^{52}{C_2}}}\]

On the back of \[52\] cards,\[26\] red colour and \[26\] black colour.

\[{\text{P}}\left( {{\text{both cards of opposite colour}}} \right) = \dfrac{{^{26}{C_1}\times {{ }^{26}}{C_1}}}{{^{52}{C_2}}} = \dfrac{{26 \times 26}}{{\dfrac{{52 \times 51}}{2}}} = \dfrac{{26}}{{51}}\]

Correct answer: C