Probability Test - 23

P(A) ; denote the probability that bus A will be late.
P(B) ; denote the probability that bus B will be late.
$$P(A)=\dfrac {1}{5}, P(B)=\dfrac {7}{25}$$
$$P\left (\dfrac {B}{A}\right )=\dfrac {9}{10}\Rightarrow P(A\cap B)=\dfrac {9}{50}$$
(i) $$P(\overline A\cap \overline B)=1-P(A\cup B)$$
$$=1-\left (\dfrac {1}{5}+\dfrac {7}{25}-\dfrac {9}{50}\right )=\dfrac {35}{50}=\dfrac {7}{10}$$
$$(ii) P\left (\dfrac {A}{B}\right )=\dfrac {P(A)\cdot P\left (\dfrac {B}{A}\right )}{P(B)}=\dfrac {\dfrac {1}{5}\times \dfrac {9}{10}}{\dfrac {7}{25}}=\dfrac {18}{28}$$

Given,In building programme the event that all materials will be delivered at the correct time is $$M$$ and event that building programme will be completed on time is $$F$$.
$$P(M)=0.8,P(M \cap F)=0.65$$$$P(F/M)=\dfrac{P(F \cap M)}{P(M)}=\dfrac{0.65}{0.8}=\dfrac{13}{16}$$If $$P(F)=0.7$$
The probability that the building programme will be completed on tim if all materials are not delivered at the correct time=$$P(F/\overline M)$$
$$\Rightarrow P(F/\overline M)=\dfrac{P(F \cap \overline M)}{P(\overline M)}=\dfrac{(P(F)-P(F \cap M))}{(1-P(M))}$$
$$\Rightarrow P(F/\overline M)=\dfrac{(0.7-0.65)}{(1-0.8)}=\dfrac{0.05}{0.2}=\dfrac{1}{4}$$

Given, S and T are two events on sample space such that,
$$P(S)=0.5,P(T)=0.69,P(S/T)=0.5$$
$$\Rightarrow P(S/T)=\frac{P(S \cap T)}){P(T)}$$
$$0.5\times 0.69=P(S \cap T)$$
$$\Rightarrow P(S \cap T)=0.345$$
$$P(S \cup T)=P(S)+P(T)-P(S \cap T)=0.5+0.69-0.345=0.845$$

Given $$P(S)=0.5,P(T)=0.69,P(S/T)=0.5$$
$$\Rightarrow P(S/T)=\frac{P(S \cap T)}{P(T)}$$
Therefore $$P(S \cap T)=P(S/T).P(T)=0.5*0.69=0.345$$
Consider $$P(S).P(T)=0.5*0.69=0.345$$
since $$P(S \cap T)=P(S).P(T)$$
$$\Rightarrow$$Events S and T are Independent

Since, total no. of balls = $$9+7+4 = 20$$
 No. of balls that are not red = $$7+4 = 11$$
$$\therefore $$ Required probability = $$\displaystyle \frac {11}{20} $$
$$\therefore $$ Option D is correct.

$$E$$$$ :$$ only one hits the target
$$\displaystyle = A\overline B\overline C \cup \overline AB\overline C\cup \overline A\overline BC$$
$$\displaystyle P(A\overline B\overline C/E)=\frac {(0.3)(0.5)(0.6)}{0.44}=\frac {0.09}{0.44}$$

$$P(\overline A \cup B) + P(A \cap \overline B)$$
$$= P(A) + P(B) \times 2P(A \cap B)$$
$$= 0.4 + 0.8 2(0.22) = 0.76$$

Let $$S$$ be the sample space
$$\therefore S=\left\{ 1,2,3,4,5,6 \right\} \times \left\{ 1,2,3,4,5,6 \right\} \\ \therefore n\left( S \right) =36$$
Let $${ E }_{ 1 }\equiv $$ the event that the sum of the numbers coming up is $$9$$
and $${ E }_{ 2 }\equiv $$ the event of occurrence of $$5$$ on the first die.
$${ E }_{ 1 }\equiv \left\{ \left( 3,6 \right) ,\left( 6,3 \right) ,\left( 4,5 \right) ,\left( 5,4 \right)  \right\} \\ \therefore n\left( { E }_{ 1 } \right) =4\\ { E }_{ 2 }\equiv \left\{ \left( 5,1 \right) ,\left( 5,2 \right) ,\left( 5,3 \right) ,\left( 5,4 \right) ,\left( 5,5 \right) ,\left( 5,6 \right)  \right\} \\ \therefore n\left( { E }_{ 2 } \right) =6\\ { E }_{ 1 }\cap { E }_{ 2 }=\left\{ \left( 5,4 \right)  \right\} \\ \therefore n\left( { E }_{ 1 }\cap { E }_{ 2 } \right) =1$$
Now $$\displaystyle P\left( { E }_{ 1 }\cap { E }_{ 2 } \right) =\frac { n\left( { E }_{ 1 }\cap { E }_{ 2 } \right)  }{ n\left( S \right)  } =\frac { 1 }{ 36 } $$ and $$\displaystyle P\left( { E }_{ 2 } \right) =\frac { n\left( { E }_{ 2 } \right)  }{ n\left( S \right)  } =\frac { 6 }{ 36 } =\frac { 1 }{ 6 } $$
$$\therefore$$ Required probability $$\displaystyle P\left( \frac { { E }_{ 1 } }{ { E }_{ 2 } }  \right) =\frac { P\left( { E }_{ 1 }\cap { E }_{ 2 } \right)  }{ P\left( { E }_{ 2 } \right)  } =\frac { \frac { 1 }{ 36 }  }{ \frac { 1 }{ 6 }  } =\frac { 1 }{ 6 } $$

Since, $$E$$ and $$F$$ are independent events with $$P(E) = 0.7$$ and $$P(F) = 0.3 $$
$$\therefore $$ P$$\displaystyle (E\cap F) = P(E)\times P(F)$$
$$\Rightarrow $$ P$$\displaystyle (E\cap F)  = 0.7\times 0.3 $$
$$\Rightarrow $$ P$$\displaystyle (E\cap F)  = 0.21$$
$$\therefore $$ Option C is correct.