Chapter 13: Probability

Ans: Let us have following notations

E: The event when A works 

F:  The event when B works

G: The event when C works

K: The event that the defective item is produced

$\therefore P\left( E \right)=0.5$ 

$\therefore P\left( F \right)=0.3$

$\therefore P\left( G \right)=0.2$

$\therefore P\left( K\left| E \right. \right)=0.01$ 

$\left( K\left| E \right. \right)$: The event that the defective item is produced from A

$\therefore P\left( K\left| F \right. \right)=0.05$

$\left( K\left| F \right. \right)$: The event that the defective item is produced from B

$\therefore P\left( K\left| G \right. \right)=0.07$

The event that the defective item is produced from C

By Baye’s theorem 

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

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{0.5\times 0.01}{0.5\times 0.01+0.3\times 0.05+0.2\times 0.07}\]

Thus the probability that that the item is produced from A and found defective $\frac{5}{34}$

Ans: Let us have following notations

E: The event when a card is diamond

F:  The event when the card is not diamond

K: The event to denote lost card 

Clearly E and F are contemporary Events

$\therefore P\left( E \right)=\frac{13}{52}$ (since $13$cards are diamonds)

$\therefore P\left( F \right)=\frac{39}{52}$

$\left( K\left| E \right. \right)$: The event of drawing two cards when one diamond card is lost

Since one diamond card is lost therefore left total cards are 

$51$ and left diamond cards are $13$

$\therefore P\left( K\left| E \right. \right)=\frac{^{12}{{C}_{2}}}{^{51}{{C}_{2}}}$

$\Rightarrow P\left( K\left| E \right. \right)=\frac{22}{425}$

$\left( K\left| F \right. \right)$: The event of drawing two cards when one non diamond card is lost

Since one card is lost therefore left total cards are $51$

$\therefore P\left( K\left| F \right. \right)=\frac{^{13}{{C}_{2}}}{^{51}{{C}_{2}}}$

$\Rightarrow P\left( K\left| F \right. \right)=\frac{26}{425}$

By Baye’s theorem 

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

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{\frac{1}{4}\times \frac{22}{425}}{\frac{1}{4}\times \frac{22}{425}+\frac{3}{4}\times \frac{26}{425}}\]

Thus the probability of the lost card being a diamond is $\frac{11}{50}$

Ans: Let us have following notations

E: The event when A speaks truth

F:  The event when A speak false

K: The event it is head 

Clearly E and F are contemporary Events

$\therefore P\left( E \right)=\frac{4}{5}$ (since $13$cards are diamonds)

$\therefore P\left( F \right)=\frac{1}{5}$

$\left( K\left| E \right. \right)$: The event it is head and he speaks truth

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

$\left( K\left| F \right. \right)$: The event it is head and he speaks false

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

By Baye’s theorem 

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

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{\frac{4}{5}\times \frac{1}{2}}{\frac{4}{5}\times \frac{1}{2}+\frac{1}{5}\times \frac{1}{2}}\]

Thus the probability of the lost card being a diamond is $\frac{4}{5}$

Ans: Given $A\subset B$

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

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

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

Now $P\left( B \right)\le 1$

$\therefore \frac{1}{P\left( B \right)}\ge 1$

Hence option C is correct, i.e  $P\left( A\left| B \right. \right)\ge P\left( A \right)$