Probability

49. It is known that the sum of all the probabilities in a probability distribution is one.

(i) Sum of the probabilities = 0.4 + 0.4 + 0.2 = 1

Therefore, the given table is a probability distribution of random variables.

(ii) It can be seen that for X = 3, P (X) = −0.1

It is known that proba

48. If $A\subset B,$ then $A\cap B=A$

$P(A\cap B)=P(A)$ also $P(A)<P(B)$

Consider,

$P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}\ne \frac{P(B)}{P(A)}$

$\Rightarrow P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}$ $\_\_\_\_\_(1)$

We know, $P(B)\le 1$ $\Rightarrow \frac{1}{P(B)}\ge 1$

From eq. $(1)$ , we have

$\frac{P(A)}{P(B)}\ge P(A)$

$\Rightarrow P(A/B)\ge P(A)$ $\_\_\_\_\_(2)$

$\Rightarrow P(A/B)$ is not less than $P(A)$

Hence, from eq. $(2)$ it can be concluded that the relation given in alternative $(C)$ is correct.

47. Let, $E_1$ and $E_2$ be the event such that

$E_1:$ $A$ speak truth

$E_2:$ $A$ speak false

$X:$ that head appears

$P(E_1)=\frac{4}{5}$ and $P(E_2)=1-P(E_1)$ = $=1-\frac{4}{5}=\frac{1}{5}$

If a coin tossed, then it may result either head $(H)$ or tail $(T)$ . The probability of getting a head is $\frac{1}{2}$ whether $A$&nbs

46. Let, $E_1:$ event of choosing a diamond card

$E_2:$ event of choosing a card which is not diamond

$A:$ denote the lost card, out of $52$ cards

We know,

$13$ cards are diamond

$39$ cards are not diamond

$P(E_1)=\frac{13}{52}=\frac{1}{4}$

$P(E_2)=\frac{39}{52}=\frac{3}{4}$

When one diamond card is lost, there are $12$ diamond cards out of $51$ card

45. Let, $E_1:$ event of time consume by $A$

$E_2:$ event of time consume by $B$

$E_3:$ event of time consume by $C$

$P(E_1)=50\mathrm{\%}=\frac{50}{100}=\frac{1}{2}$

$P(E_2)=30\mathrm{\%}=\frac{30}{100}=\frac{3}{10}$

$P(E_3)=20\mathrm{\%}=\frac{20}{100}=\frac{1}{5}$

Let, $A:$ event of producing the defective item. Therefore,

$P(A/E_1)=1\mathrm{\%}=\frac{1}{100}$

$P(A/E_2)=5\mathrm{\%}=\frac{5}{100}$

$P(A/E_3)=7\mathrm{\%}=\frac{7}{100}$

Therefore, by Baye’s theorem,

$P(E_1/A)=$ probability that the defective item was produced by $A$ ,

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

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.+\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\times \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$100$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\times \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}{\frac{1}{100}\left(\frac{1}{2}+\frac{15}{10}+\frac{7}{5}\right)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}{\frac{1}{100}\left(\frac{5+15+14}{10}\right)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$34$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}$

$=\frac{1}{2}\times \frac{10}{34}=\frac{5}{34}$

44. Let, $E_1:$ event that outcome on the die is $5$ or $6$

$E_2:$ event that outcome on the die is $1,2,3$ or $4$

$A:$ event of getting exactly one head

$P(E_1)=\frac{2}{6}=\frac{1}{3}$

$P(E_2)=\frac{4}{6}=\frac{2}{3}$

$P(A/E_1)=$ probability of getting exactly one head by tossing the coin three times if she gets $5$ or $6$ $=\frac{3}{8}$

$P(A/E_2)=$ probabilit

43. Let, $E_1:$ event that first group will win

$E_2:$ event that second group will win

$A:$ event that new product will produce

$P(E_1)=0.6=\frac{6}{10}=\frac{3}{5}$

$P(E_2)=0.4=\frac{4}{10}=\frac{2}{5}$

$P(A/E_1)=P$ (introducing new product by group $A$ ) $=0.7=\frac{7}{10}$

$P(A/E_2)=P$ ( introducing new product by group $B$ ) $=0.3=\frac{3}{10}$

Therefore, by Baye’s theorem,

$P(E_2/A)=$ probability that n

42. Let, $E_1:$ event items produced by $A$

$E_2:$ event items produced by $B$

$X:$ event that produced item was found to be defective

$P(E_1)=60\mathrm{\%}=\frac{3}{5}$

$P(E_2)=40\mathrm{\%}=\frac{2}{5}$

$P(X/E_1)=P$ (items produced by machine $A$ which is defective) $=2\mathrm{\%}=\frac{2}{100}$

$P(X/E_2)=P$ ( items produced by machine $B$ which is defective) $=1\mathrm{\%}=\frac{1}{100}$

Therefore, by Baye’s

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) $=\frac{4000}{12000}=\frac{1}{3}$

$P(E_3)=P$ (dri

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