Chapter 13: Probability

Ans: It is given that $P\left( A \right)=0.3$, $P\left( B \right)=0.6$

Where A and B are independent events

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

$\Rightarrow P\left( \text{A and B} \right)=0.3\times 0.6$

Thus we found that $P\left( \text{A and B} \right)=0.18$

Ans: It is given that $P\left( A \right)=0.3$, $P\left( B \right)=0.6$

Where A and B are independent events

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

Now we know that

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

Also we know that 

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

$\text{     }P\left( A\bigcap B' \right)=0.3-0.18$

Thus we found that $P\left( \text{A and not B} \right)=0.12$

Ans: It is given that $P\left( A \right)=0.3$, $P\left( B \right)=0.6$

Where A and B are independent events

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

$\Rightarrow P\left( \text{A and B} \right)=0.3\times 0.6$

Thus we found that $P\left( \text{A and B} \right)=0.18$

Also we know that 

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

$\text{         }P\left( A\bigcup B \right)=0.3+0.6-0.18$

Thus we know that $P\left( A\text{ or B} \right)=0.72$

Ans: It is given that $P\left( A \right)=0.3$, $P\left( B \right)=0.6$

Where A and B are independent events

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

$\Rightarrow P\left( \text{A and B} \right)=0.3\times 0.6$

Thus we found that $P\left( \text{A and B} \right)=0.18$

Also we know that 

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

$\text{         }P\left( A\bigcup B \right)=0.3+0.6-0.18$

Thus we know that $P\left( A\text{ or B} \right)=0.72$

$P\left( \text{neither A nor B} \right)=P\left( A\bigcup B \right)'$

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

$\text{              }P\left( A\bigcup B \right)'=1-0.72$

Thus $P\left( \text{neither A nor B} \right)=0.28$

Ans: It is given that a die is tossed thrice, the number of elements in sample space for each throw is $6$

Now probability of getting atleast odd once=1-probability of getting even numbers in all three events

Now we know that  

Probability of getting even numbers in all three events=${{\left( \frac{1}{2} \right)}^{3}}\text{ }$

Probability of getting at least odd once=1-$\frac{1}{8}$

Thus probability of getting at least odd once=$\frac{7}{8}$

Ans: It is given in the question that 

Number of red balls =8 and number of black balls =10

Now probability of getting  red ball at first draw,say P(A)=$\frac{8}{18}$=$\frac{4}{9}$

Also probability of getting  red ball at second draw, say (P)=$\frac{8}{18}$=$\frac{4}{9}$

$\therefore P\left( A\bigcap B \right)=\frac{4}{9}\times \frac{4}{9}$(since the events are independent)

$=\frac{16}{81}$

The probability of getting both balls red is $\frac{16}{81}$

Ans: It is given in the question that 

Number of red balls =8 and number of black balls =10

Now probability of getting  black ball at first draw=$\frac{10}{18}$=$\frac{5}{9}$

Also probability of getting  red ball at second draw=$\frac{8}{18}$=$\frac{4}{9}$

Also above two events are independent therefore their intersection is given by

$=\frac{5}{9}\times \frac{4}{9}$

$=\frac{20}{81}$

$\begin{align} & \text{probability of getting black ball in first draw} \\ & \text{and red ball in second draw =}\frac{20}{81} \\ \end{align}$

Ans: Let us have following notations 

A: first ball is black and the other is red

Now probability of getting a black ball in first draw, $P\left( A \right)=\frac{10}{18}$

$=\frac{5}{9}$

As the ball is replaced after first throw, probability of getting a red ball in second draw $=\frac{8}{18}$

$=\frac{4}{9}$

Probability of getting first ball as black and second ball as red $=\frac{5}{9}\times \frac{4}{9}$

$=\frac{20}{81}$

Probability of drawing red ball in first draw $=\frac{8}{18}$

$=\frac{4}{9}$

Probability of getting black ball in second draw$=\frac{10}{81}$

$=\frac{5}{9}$

Probability of drawing first ball as red and second as black $=\frac{5}{9}\times \frac{4}{9}$

$=\frac{20}{81}$

Probability of getting one of them is black and the other is red$=\frac{20}{81}+\frac{20}{81}$

$=\frac{40}{81}$

Probability of getting one of the ball is black and the other is red is $\frac{40}{81}$

Ans: Given that the two events A and B are independent with probabilities given as shown

$P\left( A \right)=\frac{1}{2}$ and $P\left( B \right)=\frac{1}{3}$

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

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

Probability of problem being solved is given by 

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

\[\Rightarrow P\left( A\bigcup B \right)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\]

Thus probability of problem being solved is $\frac{2}{3}$

Ans: Given that the two events A and B are independent with probabilities given as shown

$P\left( A \right)=\frac{1}{2}$ and $P\left( B \right)=\frac{1}{3}$

Either one of them solved the answer. So either A solve the answer but not B or B solved the answer but not A

$P\left( A \right).P\left( B' \right)+P\left( A' \right).P\left( B \right)$

$=\frac{1}{2}\left( \frac{2}{3} \right)+\frac{1}{2}\left( \frac{1}{3} \right)$

$=\frac{2}{6}+\frac{1}{6}$

$=\frac{3}{6}$

$=\frac{1}{2}$

Probability of one of them solves the answer is $\frac{1}{2}$

Ans: Given a deck of $52$ cards 

We know that $13$cards are spades and $4$are aces 

$P\left( E \right)=\frac{13}{52}$ and $P\left( F \right)=\frac{4}{52}$

Also in the deck only $1$ card is ace of spade

I.e $P\left( E\bigcap F \right)=\frac{1}{52}$

It is clearly visible that 

$P\left( E \right)P\left( F \right)=\frac{13}{52}\times \frac{4}{52}$ equals $P\left( E\bigcap F \right)$

Hence the events are independent

(ii)E: ‘the card drawn is black’

 F: ‘the card drawn is a king’

Ans:  We know that in the deck of cards $26$ cards are black and $4$ cards are kings

$P\left( E \right)=\frac{26}{52}$ and $P\left( F \right)=\frac{4}{52}$

Also there are 2 cards which are black as well as king

i.e $P\left( E\bigcap F \right)=\frac{2}{52}$

It is clearly visible that $\frac{26}{52}\times \frac{4}{52}=\frac{2}{52}$

Hence these are independent events

(iii) E: ‘the card drawn is a king and queen’

 F: ‘the card drawn is a queen or jack’

Ans: In a deck of $52$ cards each king, queen and jack is $4$

$\therefore P\left( E \right)=P\left( F \right)=\frac{8}{52}$

There are $4$cards which which are king and queen or jack

Hence its probability is $\frac{4}{52}$

Clearly E and F are not independent 

Ans: Let us have following notations 

H: Students who read Hindi newspapers

E: Students who read English newspapers

$\therefore P\left( H \right)=0.6$ and

$\therefore P\left( E \right)=0.4$

Also it is given that 

$P\left( H\bigcap E \right)=0.2$

Therefore the required probability is given by

\[P\left( H\bigcup E' \right)=1-P\left( H\bigcup E \right)\]

\[P\left( H\bigcup E' \right)=1-\left( P\left( H \right)+P\left( E \right)-P\left( H\bigcap E \right) \right)\]

\[P\left( H\bigcup E' \right)=1-0.8\]

Hence the probability that she reads neither Hindi nor English is $0.2$

1. If she reads Hindi news paper, find the probability that she reads English news paper.

Ans: Let us have following notations 

H: Students who read Hindi newspapers

E: Students who read English newspapers

$\therefore P\left( H \right)=0.6$ and

$\therefore P\left( E \right)=0.4$

Also it is given that 

$P\left( H\bigcap E \right)=0.2$

Now required probability is given by 

\[\text{    }P\left( E\left| H \right. \right)=\frac{P\left( H\bigcap E \right)}{P\left( H \right)}\]

\[\Rightarrow P\left( E\left| H \right. \right)=\frac{0.2}{0.6}\]

Thus the probability that she reads English newspapers if she reads Hindi news papers is $\frac{1}{3}$

1. If she reads English newspapers, find the probability that she reads Hindi newspapers.

Ans: Let us have following notations 

H: Students who read Hindi newspapers

E: Students who read English newspapers

$\therefore P\left( H \right)=0.6$ and

$\therefore P\left( E \right)=0.4$

Also it is given that 

$P\left( H\bigcap E \right)=0.2$

The required probability is given by

$\text{   }P\left( E\left| H \right. \right)=\frac{P\left( H\bigcap E \right)}{P\left( H \right)}$

$\Rightarrow P\left( E\left| H \right. \right)=\frac{0.2}{0.4}$

Therefore the probability that she reads Hindi newspapers provided she reads English newspapers is $\frac{1}{2}$

Ans: Out of six numbers in a die, the only even prime number is $2$

Therefore probability of getting an even prime number is given by

Say $P\left( E \right)=\frac{1}{36}$

Hence probability of getting an even prime number is $\frac{1}{36}$

Ans: Let $P\left( A \right)=p$ and $P\left( B \right)=q$ $0<p,q<1$

It is given that they are mutually exclusive 

$\therefore P\left( A\bigcap B \right)=0$

But $P\left( A \right)P\left( B \right)=pq$ which is not necessarily zero

Hence they are not independent

1. \[P\left( A'B' \right)=\left[ 1-P\left( A \right) \right]\left[ 1-P\left( B \right) \right]\]

Ans: Two events are said to be independent if $P\left( AB \right)=P\left( A \right)P\left( B \right)$

Let us solve for option B first

$P\left( A'B' \right)=\left[ 1-P\left( A \right) \right]\left[ 1-P\left( B \right) \right]$

$\Rightarrow P\left( A'\bigcap B' \right)=1-P\left( A \right)-P\left( B \right)+P\left( A \right)P\left( B \right)$

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

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

Hence we found that $P\left( AB \right)=P\left( A \right)P\left( B \right)$

Hence these are independent events

1. \[P\left( A \right)=P\left( B \right)\]

Ans: If  $P\left( A \right)=P\left( B \right)$ then

Each of them has probability of $\frac{1}{2}$ 

Surely it can be sometime true but not a sufficient condition for two events to be independent

1. \[P\left( A \right)+P\left( B \right)=1\]

Ans: Similarly If  $P\left( A \right)+P\left( B \right)=1$ it is not necessary condition for two events to be independent