Probability

41. Let, $E_1:$ event that the driver is scooter driver

$E_2:$ event that the driver is car driver

$E_3:$ event that the driver is truck driver

$A:$ event the person meets with accident

Total number of drivers $=2000+4000+6000=12000$

$P(E_1)=P$ (driver is a scooter driver) $=\frac{2000}{12000}=\frac{1}{6}$

$P(E_2)=P$ (driver is a car driver) $=\frac{4000}{12000}=\frac{1}{3}$

$P(E_3)=P$ (driver is a truck driver) $=\frac{6000}{12000}=\frac{1}{2}$

$P(A/E_1)=P$ (scooter driver met with an accident) $=0.01=\frac{1}{100}$

$P(A/E_2)=P$ (car driver met with an accident) $=0.03=\frac{3}{100}$

$P(A/E_3)=P$ (truck driver met with an accident) $=0.15=\frac{15}{100}$

The probability that the driver is scooter driver, given that he met with an accident is given by $P(E_1/A)$

By Baye's theorem,

$P(E_1/A)=\frac{P(E_1).P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)+P(E_3).P(A/E_3)}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.*\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.*\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.*\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$100$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.*\raisebox{1ex}{$15$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$600$}\right.}{\frac{1}{100}\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.+1+\raisebox{1ex}{$15$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$600$}\right.}{\frac{1}{100}\left(\frac{1+6+45}{6}\right)}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$600$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.*\raisebox{1ex}{$52$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}$

$=\frac{1}{600}*\frac{600}{52}=\frac{1}{52}$

40. Let, $E_1:$ event of choosing $Q$ headed coin

$E_2:$ event of choosing biased coin

$E_3:$ event of choosing unbiased coin

$P(E_1)=P(E_2)=P(E_3)=\frac{1}{3}$

Let $A$ be the event that shows head,

$P(A/E_1)=P$ (coin shows heads, given that it is a headed coin) $=1$

$P(A/E_2)=P$ (coin showing up head, given that it is biased coin) $=\frac{75}{100}=\frac{3}{4}$

$P(A/E_3)=P$ (coin showing head, given that it is unbiased coin) $=\frac{1}{2}$

The probability that the coin is two headed, given that it shows heads, is given by $P(E_1/A)$

Using Baye's theorem,

$P(E_1/A)=\frac{P(E_1).P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)+P(E_3).P(A/E_3)}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.*1}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.*1+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.*\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.*\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\frac{4+3+2}{12}}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$12$}\right.}=\frac{1}{3}*\frac{12}{9}=\frac{4}{9}$

39. Let, $E_1:$ event person has disease

$E_2:$ event person has no disease

$A:$ be event that blood test is positive

As $E_1$ and $E_2$ are events which are complementary to each other,

Then, $P(E_1)+P(E_2)=1$

$P(E_2)=1-P(E_1)$

$P(E_1)=0.1\mathrm{\%}=\frac{0.1}{100}=0.001$

$\therefore$ $P(E_2)=1-0.001$

$=0.999$

$P(A/E_1)=P$ (result is positive given that person has disease) $=99\mathrm{\%}=0.99$

$P(A/E_2)=P$ (result is positive given that person has no disease) $=0.5\mathrm{\%}=0.005$

Now, the probability that person has a disease, given that his test result is positive is $P(E_1/A)$

By using Baye's theorem,

$P(E_1/A)=\frac{P(E_1).P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)}$

$=\frac{0.001*0.99}{0.001*0.99+0.999*0.005}$

$=\frac{0.00099}{0.00099+0.004995}$

$=\frac{0.00099}{0.005985}=\frac{990}{5985}=\frac{110}{665}$

$\Rightarrow P(E_1/A)=\frac{22}{133}$

38. Let, $E_1$ :event that knows the answer

$E_2$ :event that he guesses the answer

$P(E_1)=\frac{3}{4}$ and $P(E_2)=\frac{1}{4}$

Let $A$ be the event that the answer is correct

Also, $P(A/E_1)=P$ (correct answer given that he knows) $=1$

$P(A/E_2)=P$ (correct answer given that he guess) $=\frac{1}{4}$

Now, probability that he knows the answer given that he answer it correctly is $P(E_1/A)$ ,

By Baye's theorem,

$P(E_1/A)=\frac{P(E_1).P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)}$

$=\frac{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.*1}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.*1+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.*\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=\frac{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$16$}\right.}$

$=\frac{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\frac{12+1}{16}}=\frac{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\raisebox{1ex}{$13$}\!\left/ \!\raisebox{-1ex}{$16$}\right.}$

$=\frac{3}{4}*\frac{16}{13}=\frac{12}{13}$

$P(E_1/A)=\frac{12}{13}$

37. Let, $E_1$ :event that the student in hostel

$E_2$ : event that the student is day scholar

$A$ : be the event of the chosen student get grade $A$

$P(E_1)=60\mathrm{\%}=\frac{60}{100}=0.6$

$P(E_2)=40\mathrm{\%}=\frac{40}{100}=0.4$

$P(A/E_1)=P$ (student getting an $A$ grade is hostler) $=30\mathrm{\%}$ $=\frac{30}{100}=0.3$

$P(A/E_2)=P$ ( student getting an $A$ grade is a day scholar) $=20\mathrm{\%}=\frac{20}{100}=0.2$

Then, by Baye's theorem,

$P(E_1/A)=\frac{P(E_1).P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)}$

$=\frac{0.6*0.3}{0.6*0.3+0.4*0.2}=\frac{0.18}{0.26}=\frac{18}{26}=\frac{9}{13}$

36. Let $E_1$ be the event of selecting first bag

$E_2$ be the event of selecting second bag

$\therefore P(E_1)=P(E_2)=\frac{1}{2}$

Let $A$ be the event of getting a red ball

$P(A/E_1)=P$ (drawing a red ball from first bag)

$=\frac{4}{8}=\frac{1}{2}$

$P(A/E_2)=P$ (drawing a red ball from second bag)

$=\frac{2}{8}=\frac{1}{4}$

Then, probability of drawing a ball from the first bag which is red, is given by Baye's theorem,

$P(E_1/A)=\frac{P(E_1).P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.*\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.*\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.*\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}=\frac{2}{3}$

35. Number of balls contain in win $=5$ red

$=5$ black

Let the red ball be drawn in the first attempt

$\therefore$ $P$ (drawing a red ball) $=\frac{5}{10}=\frac{1}{2}$

By question, if added two red balls to the win, then $7$ red balls and $5$ black balls contain.

$P$ (drawing a red ball) $=\frac{7}{12}$

Let a black ball be drawn at first attempt

$\therefore$ $P$ (drawing a black ball) $=\frac{5}{10}=\frac{1}{2}$

If two black balls are added to the win, then $7$ black balls and $5$ red balls contain.

$P$ (drawing a red ball) $=\frac{5}{12}$

Therefore, probability of drawing the second ball as of red colour is

$=\left(\frac{1}{2}*\frac{7}{12}\right)+\left(\frac{1}{2}*\frac{5}{12}\right)=\frac{1}{2}\left(\frac{7}{12}+\frac{5}{12}\right)=\frac{1}{2}*\frac{12}{12}=\frac{1}{2}*1=\frac{1}{2}$

$$34. $P(A^{\prime }{B}^{\prime })=[1-P(A)][1-P(B)]$

$\Rightarrow P(A^{\prime }\cap B^{\prime })=1-P(A)-P(B)+P(A)P(B)$

$\Rightarrow -[P(A)+P(B)-P(A\cap B)]=-P(A)-P(B)+P(A)(B)$

$\Rightarrow -P(A)-P(B)+P(A\cap B)=-P(A)-P(B)+P(A)P(B)$

$\Rightarrow P(A\cap B)=P(A).P(B)$

Therefore, it shows A and B are independant

33. The outcome of rolling $2$ dice is $36$ . The only prime number is $2$ .

Let,  $E=$ event of getting an even prime number on each die

$E=\{(2,2)\}$ therefore

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

Therefore, the correct answer is $(D)$ $\frac{1}{36}$

32. Let,

$P(H)=$ denote the student who read Hindi newspaper

$P(E)=$ denote the student who read English newspaper

Then, $P(H)=60\mathrm{\%}=\frac{6}{10}=\frac{3}{5}$

$P(E)=40\mathrm{\%}=\frac{4}{10}=\frac{2}{5}$

$\mathrm{i.\; P}(H\cap E)=20\mathrm{\%}=\frac{2}{10}=\frac{1}{5}$

$\Rightarrow P(H\cup E{)}^{\prime }=1-[P(H)+P(E)-P(H\cap E)]$

$=1-\left[\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\right]$

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

$=1-\frac{4}{5}=\frac{1}{5}$

ii. Probability of randomly chosen student that reads English newspaper, if she reads Hindi newspaper, is given by

$P(E/H)=\frac{P(E\cap H)}{P(H)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}=\frac{1}{3}$

iii. Probability that randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by

$P(H/E)=\frac{P(H\cap E)}{P(E)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}=\frac{1}{2}$