Chapter 13: Probability

Ans: It is given in the question that $P\left( A \right)=\frac{3}{5}$ and $P\left( B \right)=\frac{1}{5}$

Also it is given that A and B are independent events

$\text{ }\therefore P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( A\bigcap B \right)=\frac{3}{5}\times \frac{1}{5}$

Thus $P\left( A\bigcap B \right)=\frac{3}{25}$

Ans: Given a deck of  $52$ cards 

We know that the number of black cards are $26$

Let us have a notation $P\left( A \right)$ denote the probability of picking a black card at first chance

$\therefore P\left( A \right)=\frac{26}{52}$

Similarly, let us have a notation $P\left( B \right)$ denote the probability of picking a black card at second chance.

Since there is no replacement 

$\therefore P\left( B \right)=\frac{25}{51}$

The probability that the both the cards are black is

$P\left( A\bigcap B \right)=\frac{1}{2}\times \frac{25}{51}$

$=\frac{25}{102}$

The probability that the both the cards are black is $\frac{25}{102}$

Ans: Let A,B and C be the events and their probabilities defined as 

A: the first orange is good

$\therefore P\left( A \right)=\frac{12}{15}$

B: the second orange is good

$\therefore P\left( B \right)=\frac{11}{14}$

C: the third orange is good

$\therefore P\left( C \right)=\frac{10}{13}$

It is given that the box is approved for sale only when all the oranges are good.

Probability that the box is approved for sale=probability of all oranges to be good

Therefore probability of all oranges to be good=$\frac{12}{15}\times \frac{11}{14}\times \frac{10}{13}$

Thus probability that the box is approved for sale=$\frac{44}{91}$=0.48

Ans: The sample space is given by

$S=\left\{ \begin{align} & \left( H,1 \right),\left( H,2 \right)......\left( H,6 \right) \\ & \left( T,1 \right),\left( T,2 \right)........\left( T,6 \right) \\ \end{align} \right\}$

According to the question

$A=\left\{ \left( H,1 \right),\left( H,2 \right)......\left( H,6 \right) \right\}$

$\therefore P\left( A \right)=\frac{6}{12}$

\[B=\left\{ \left( H,3 \right),\left( T,3 \right) \right\}\]

$\therefore P\left( B \right)=\frac{2}{12}$

Also $\left( A\bigcap B \right)=\left\{ \left( H,3 \right) \right\}$

$\therefore P\left( A\bigcap B \right)=\frac{1}{12}$

We know that for A and B to be independent events

$P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

Now $P\left( A \right)P\left( B \right)=\frac{6}{12}\times \frac{2}{12}$

$\Rightarrow P\left( A \right)P\left( B \right)=\frac{1}{12}$

Which is as same as $P\left( A\bigcap B \right)$

Thus A and B are independent events

Ans: The sample space is given by

$S=\left\{ 1,2,3,4,5,6 \right\}$

According to the question

$A=\left\{ 2,4,6 \right\}$

$\therefore P\left( A \right)=\frac{3}{6}$

\[B=\left\{ 1,2,3 \right\}\]

$\therefore P\left( B \right)=\frac{3}{6}$

Also $\left( A\bigcap B \right)=\left\{ 2 \right\}$

$\therefore P\left( A\bigcap B \right)=\frac{1}{6}$

We know that for A and B to be independent events

$P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

Now $P\left( A \right)P\left( B \right)=\frac{3}{6}\times \frac{3}{6}$

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

Which is not  as same as $P\left( A\bigcap B \right)$

Thus A and B are not independent events

Ans: Given $P\left( E \right)=\frac{3}{5}$,  $P\left( F \right)=\frac{3}{10}$ and $P\left( E\bigcap F \right)=\frac{1}{5}$

We know that for A and B to be independent events

$P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

Now $P\left( A \right)P\left( B \right)=\frac{3}{5}\times \frac{3}{10}$

$\Rightarrow P\left( A \right)P\left( B \right)=\frac{9}{50}$

Also $P\left( E\bigcap F \right)=\frac{1}{5}$

Thus $P\left( A \right)P\left( B \right)\ne P\left( E\bigcap F \right)$

And hence A and B are not independent events

Ans: It is given in the question that $P\left( A \right)=\frac{1}{2}$, $P\left( B \right)=p$ and $P\left( A\bigcup B \right)=\frac{3}{5}$

We know that if two events A and B are mutually exclusive then $P\left( A\bigcap B \right)=0$

Also we know that 

$P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$

$\Rightarrow \frac{3}{5}=\frac{1}{2}+p-0$

$\Rightarrow p=\frac{3}{5}-\frac{1}{2}$

$\Rightarrow p=\frac{1}{10}$

Ans: It is given in the question that $P\left( A \right)=\frac{1}{2}$, $P\left( B \right)=p$ and $P\left( A\bigcup B \right)=\frac{3}{5}$

We know that if two events A and B are independent then $P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

Also we know that 

$P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$

$\Rightarrow \frac{3}{5}=\frac{1}{2}+p-\frac{p}{2}$ (since $P\left( A \right)P\left( B \right)=\frac{p}{2}$)

$\Rightarrow \frac{p}{2}=\frac{3}{5}-\frac{1}{2}$

$\Rightarrow p=\frac{2}{10}$

$=\frac{1}{5}$

The value of  $p$ when A and B are independent event is $\frac{1}{5}$

Ans: It is given that A and B are independent events with given probabilities as shown

$P\left( A \right)=0.3$

$P\left( B \right)=0.4$

We know that if A and B are independent events then 

$\text{    }P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( A\bigcap B \right)=0.3\times 0.4$

$\Rightarrow P\left( A\bigcap B \right)=0.12$

Ans: It is given that A and B are independent events with given probabilities as shown

$P\left( A \right)=0.3$

$P\left( B \right)=0.4$

We know that if A and B are independent events then 

$\text{    }P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( A\bigcap B \right)=0.3\times 0.4$

$\Rightarrow P\left( A\bigcap B \right)=0.12$

Also we know that 

$P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$

$\Rightarrow P\left( A\bigcup B \right)=0.3+0.4-0.12$

$\Rightarrow P\left( A\bigcup B \right)=0.58$

Ans: It is given that A and B are independent events with given probabilities as shown

$P\left( A \right)=0.3$

$P\left( B \right)=0.4$

We know that if A and B are independent events then 

$\text{    }P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( A\bigcap B \right)=0.3\times 0.4$

$\Rightarrow P\left( A\bigcap B \right)=0.12$

Also we know that $P\left( A\left| B \right. \right)$ is given by

$P\left( A\left| B \right. \right)=\frac{\left( P\left( A\bigcap B \right) \right)}{P\left( B \right)}$

$\Rightarrow P\left( A\left| B \right. \right)=\frac{0.12}{0.4}$

$=0.3$

Thus we found that $P\left( A\left| B \right. \right)=0.3$

Ans: It is given that A and B are independent events with given probabilities as shown

$P\left( A \right)=0.3$

$P\left( B \right)=0.4$

We know that if A and B are independent events then 

$\text{    }P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$

$\Rightarrow P\left( A\bigcap B \right)=0.3\times 0.4$

$\Rightarrow P\left( A\bigcap B \right)=0.12$

Also we know that \[P\left( B\left| A \right. \right)\] is given by

$P\left( B\left| A \right. \right)=\frac{\left( P\left( A\bigcap B \right) \right)}{P\left( A \right)}$

$\Rightarrow P\left( B\left| A \right. \right)=\frac{0.12}{0.3}$

$=0.4$

Thus we found that $P\left( B\left| A \right. \right)=0.4$

Ans: It is given that two events A and b have probabilities as shown

$P\left( A \right)=\frac{1}{4}$, $P\left( B \right)=\frac{1}{2}$ and \[P\left( A\bigcap B \right)=\frac{1}{8}\]

$P\left( \text{not A and not B} \right)=P\left( A'\bigcap B' \right)$

And we know that 

$\text{             }P\left( A'\bigcap B' \right)=P\left( A\bigcup B \right)'$

$\text{             }P\left( A'\bigcap B' \right)=1-P\left( A\bigcup B \right)$

Also $\text{         }P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$

$\text{          }\Rightarrow P\left( A\bigcup B \right)=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}$

\[\text{          }\Rightarrow P\left( A\bigcup B \right)=\frac{5}{8}\]

$\therefore P\left( A'\bigcap B' \right)=1-\frac{5}{8}$

Thus $P\left( \text{not A and not B} \right)=\frac{3}{8}$

Ans: It is given that $P\left( A \right)=\frac{1}{2}$, \[P\left( B \right)=\frac{7}{12}\] and \[P\left( A'\bigcup B' \right)=\frac{1}{4}\]

According to the question 

$1-P\left( A\bigcap B \right)=\frac{1}{4}$

$\Rightarrow P\left( A\bigcap B \right)=\frac{3}{4}$

We know that A and B are independent events if 

$P\left( A \right)P\left( B \right)=P\left( A\bigcap B \right)$

$\therefore P\left( A \right)P\left( B \right)=\frac{1}{2}\times \frac{7}{12}$ 

Clearly $P\left( A \right)P\left( B \right)\ne P\left( A\bigcap B \right)$

Thus A and B are not independent events