Probability

51. A coin is tossed six times and X represents the difference between the number of heads and the number of tails.

∴ X (6 H, 0T) = |6 - 0| = 6

X (5 H, 1 T)  = |5 - 1| = 4

X (4 H, 2 T)  = |4 - 2| = 2

X (3 H, 3 T)  = 3 - 3| = 0

X (2 H, 4 T)  = |2 - 4| = 2

X (1 H, 5 T)  = |1 - 5| = 4

X (0H, 6 T) = | 0 - 6| = 6

Thus, the possible values of X are 6, 4, 2, and 0.

50. There two balls may be selected as BR, RB, BR, BB, where R represents red ball and B represents black ball.

Variable X has the value 0, 1, 2, i.e., there may be no black ball, may be one black ball or both the balls are black.

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 $A$ speak truth or not.

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

The probability that there are actually a head is given by $P(E_1/X)$

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

$=\frac{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.*\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.*\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.*\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$=\frac{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.*\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\frac{1}{2}\left(\frac{4}{5}+\frac{1}{5}\right)}=\frac{4}{5}$

Option $(A)$ is correct $\frac{4}{5}$

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

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

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

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 $6$ $=\frac{3}{8}$

$P(A/E_2)=$ probabilit7y of getting exactly one head by tossing the coin three times if she gets $1,2,3$ or $4$ $=\frac{1}{2}$

Therefore, by Baye's theorem,

$P(E_2/A)=$ probability that the girl threw $1,2,3$ or $4$ with die, if she obtained exactly one head,

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

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

$=\frac{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$24$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\frac{3+8}{24}}$

$=\frac{1}{3}*\frac{24}{10}=\frac{8}{11}$

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 new product introduced was produced by second group

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

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

$=\frac{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$50$}\right.}{\raisebox{1ex}{$21$}\!\left/ \!\raisebox{-1ex}{$50$}\right.+\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$50$}\right.}$

$=\frac{6}{50}*\frac{50}{27}=\frac{2}{9}$

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 theorem,

$P(E_2/X)=$ probability that the randomly selected item was from machine $B$ ,which is defective,

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

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

$=\frac{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$500$}\right.}{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$500$}\right.+\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$500$}\right.}$

$=\frac{2}{500}*\frac{500}{8}=\frac{1}{4}$