Probability

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 positiv

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 an

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 Ba

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

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

$$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

31. i. In a deck of $52$ cards,

$13$ are spades and $4$ are aces.

$\therefore P(E)=P$ (the card drawn is spade) $=\frac{13}{52}=\frac{1}{4}$

$\therefore P(F)=P$ (the card drawn is an ace) $=\frac{4}{52}=\frac{1}{13}$

$\therefore P(E\cap F)=P$ (the card drawn is spade and an ace) $=\frac{1}{52}$

$P(E).P(F)=\frac{1}{4}\times \frac{1}{13}=\frac{1}{52}=P(E\cap F)$

$\Rightarrow P(E).P(F)=P(E\cap F)$

$E$ and $F$ are independent.

ii. In a deck of $52$ cards,

$26$ cards are black and $4$ a

30. Let, $P(A)=$ probability of solving problem by $A=\frac{1}{2}$

$P(B)=$ probability of solving problem by $B=\frac{1}{3}$

Since, the problem is solved independently by $A$ and $B$ , then,

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

$\Rightarrow P(A\cap B)=\frac{1}{2}\times \frac{1}{3}$

$=\frac{1}{6}$

$P(A^{\prime })=1-P(A)=1-\frac{1}{2}=\frac{1}{2}$

$P(B^{\prime })=1-P(B)=1-\frac{1}{3}=\frac{2}{3}$

i. Probability of the problem is solved

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

$\Rightarrow P(A\cup B)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$

$=\frac{3+2-1}{6}$

$=\frac{4}{6}=\frac{2}{3}$

ii. Probability that exactly one of the solved problem is given by $P(A)P(B^{\prime })+P(B)P(A^{\prime })=\frac{1}{2}\times \frac{2}{3}+\frac{1}{2}\times \frac{1}{3}$

$=\frac{2}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$