Chapter 13: Probability

Ans:  We have

Probability of getting a six is $p=\frac{1}{6}$

Probability of getting no six is $q=\frac{5}{6}$

There are three possible ways to get a six in three trials:

1. He gets a six only at first throw  

probability $=\frac{1}{6}$

Rupee won $=1$

1. He gets a six at second throw 

probability $=\frac{1}{6}\times \frac{5}{6}$

Rupee won $=-1+1=0$

1. He gets a six at third trial

probability $=\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}=\frac{25}{216}$

Rupee won $=-1-1+1=-1$

Thus Expected value can be calculated as shown

$1\times \frac{1}{6}+0\times \frac{5}{36}+-1\times \frac{25}{216}=\frac{11}{216}$

Ans:  Let us have following notations

 R be the event of drawing red marble 

E be the event of selecting the box A

F be the event of selecting the box B

G be the event of selecting the box C

Total numbers of marbles is $40$

Number of red marbles is $15$

$\therefore P\left( R \right)=\frac{15}{40}$

Number of red marbles in box A i.e $n\left( R\bigcap E \right)=1$

Number of red marbles in box B i.e $n\left( R\bigcap F \right)=6$

Number of red marbles in box C i.e $n\left( R\bigcap G \right)=8$

Now the probability that red marble is picked from box A is given by

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

$\Rightarrow P\left( R\left| E \right. \right)=\frac{\frac{1}{40}}{\frac{15}{40}}$

Hence E be the event of selecting the box A is $\frac{1}{15}$

Similarly, probability that red marble is picked from box B is given by

$P\left( R\left| F \right. \right)=\frac{\frac{6}{40}}{\frac{15}{40}}$

Hence probability that red marble is picked from box B is $\frac{2}{5}$

Similarly, probability that red marble is picked from box C is given by

$P\left( R\left| G \right. \right)=\frac{\frac{8}{40}}{\frac{15}{40}}$

Hence probability that red marble is picked from box C is $\frac{8}{15}$

Ans: Let us have following notations

E: The events when the person took yoga and meditation courses

$\therefore P\left( E \right)=\frac{1}{2}$

F:The events when the person took drugs

$\therefore P\left( F \right)=\frac{1}{2}$

G: the person suffered heart attack

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

From the question also we have

$P\left( G\left| E \right. \right)=0.4\times 0.7=0.28$

$P\left( G\left| F \right. \right)=0.4\times 0.75=0.30$

Now probability that found person has heart attack despite of having yoga and meditation courses is given by

$P\left( E\left| G \right. \right)=\frac{P\left( E \right)\times P\left( G\left| E \right. \right)}{P\left( E \right)\times P\left( G\left| E \right. \right)+P\left( F \right)\times P\left( G\left| F \right. \right)}$

$P\left( E\left| G \right. \right)=\frac{0.5\times 0.28}{P0.5\times 0.28+0.5\times 0.3}$

Hence probability that found person has heart attack despite of having yoga and meditation courses is $\frac{14}{29}$

Ans: It is clear that total number of determinants of second order entries being $0's\text{ or 1 }\!\!'\!\!\text{ s}$ is ${{2}^{4}}$

The value of determinant is positive for cases as shown

$\left. \left| \begin{matrix} 1 & 0  \\ 0 & 1  \\ \end{matrix} \right. \right|,\left. \left| \begin{matrix} 1 & 1  \\ 0 & 1  \\ \end{matrix} \right. \right|,\left. \left| \begin{matrix} 1 & 0  \\ 1 & 1  \\ \end{matrix} \right. \right|$

i.e favourable case is $3$

Hence probability that value of determinant is positive is $\frac{3}{16}$.

Ans: Let us have following notations

E: A fails

F: B fails

Given in the question 

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

$P\left( E\bigcap F \right)=0.15$

$P\left( E'\bigcap F \right)=0.15$

We know that 

$P\left( E'\bigcap F \right)=P\left( E \right)-P\left( E\bigcap F \right)$

$\Rightarrow P\left( E \right)=0.3$

Now the probability that A fails given B has failed is given by

$P\left( E\left| F \right. \right)=\frac{0.15}{0.3}$ 

Hence the probability that A fails given B has failed is $0.5$

Ans: Probability that A fails alone is given by

 $P\left( E\bigcap F' \right)=P\left( F \right)-P\left( E\bigcap F \right)$

$P\left( E\bigcap F' \right)=0.05$

Hence Probability that A fails alone is $0.05$

Ans:  Let us have following notations that 

E: red ball is transferred

$P\left( E \right)=\frac{3}{7}$

F: black ball is transferred

$P\left( F \right)=\frac{4}{7}$

G: red ball is drawn

When a red ball is transferred

$P\left( G\left| E \right. \right)=\frac{5}{10}$

Similarly, When a black ball is transferred

$P\left( G\left| F \right. \right)=\frac{4}{10}$

$P\left( F\left| G \right. \right)=\frac{\frac{4}{7}\times \frac{4}{10}}{\frac{4}{7}\times \frac{4}{10}+\frac{3}{7}\times \frac{5}{10}}$

Hence probability that the transferred ball is black is $\frac{16}{31}$

Ans: Given in the question $P\left( A \right)\ne 0$ and $P\left( B\left| A \right. \right)=1$

And we know that 

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

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

Hence $A\subset B$

Ans: Given $P\left( A\left| B \right. \right)>P\left( A \right)$

And we know that 

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

$\Rightarrow \frac{P\left( A\bigcap B \right)}{P\left( B \right)}>P\left( A \right)$

Hence we found that $P\left( B\left| A \right. \right)>P\left( B \right)$

Ans: Given in the question we have 

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

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

Hence 

$P\left( A\left| B \right. \right)=1$