Multiplication Theorem on Probability Questions in English

(A) Let event $E_1$ be the event that there is a lockdown and $E_2$ be the event that there is no lockdown. Let $A$ be the event that the pandemic is controlled in one month.
Given probabilities are:
$P(E_1) = 0.7$
$P(E_2) = 1 - P(E_1) = 1 - 0.7 = 0.3$
$P(A \mid E_1) = 0.8$
$P(A \mid E_2) = 0.3$
Using the Law of Total Probability:
$P(A) = P(E_1) \times P(A \mid E_1) + P(E_2) \times P(A \mid E_2)$
$P(A) = (0.7 \times 0.8) + (0.3 \times 0.3)$
$P(A) = 0.56 + 0.09 = 0.65$
Thus,the probability that the pandemic will be controlled in one month is $0.65$.

(D) Let $P(B_1)$ and $P(B_2)$ be the probabilities of selecting bag $B_1$ and bag $B_2$ respectively. Since a bag is chosen at random,$P(B_1) = P(B_2) = \frac{1}{2}$.
Let $W$ be the event of drawing a white ball.
The probability of drawing a white ball from bag $B_1$ is $P(W|B_1) = \frac{4}{4+2} = \frac{4}{6} = \frac{2}{3}$.
The probability of drawing a white ball from bag $B_2$ is $P(W|B_2) = \frac{3}{3+4} = \frac{3}{7}$.
Using the law of total probability:
$P(W) = P(B_1) \times P(W|B_1) + P(B_2) \times P(W|B_2)$
$P(W) = \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{3}{7}$
$P(W) = \frac{1}{3} + \frac{3}{14} = \frac{14+9}{42} = \frac{23}{42}$.

(B) Total number of balls $= 10$.
Number of red balls $= 6$.
We draw $3$ balls one after the other without replacement.
The probability of drawing the first red ball is $P(R_1) = \frac{6}{10}$.
The probability of drawing the second red ball given the first was red is $P(R_2|R_1) = \frac{5}{9}$.
The probability of drawing the third red ball given the first two were red is $P(R_3|R_1 \cap R_2) = \frac{4}{8}$.
The required probability is $P(R_1 \cap R_2 \cap R_3) = \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} = \frac{120}{720} = \frac{1}{6}$.

(D) Given that $A$ and $B$ are independent events,we have $P(A \cap B) = P(A) \cdot P(B)$.
We know the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $0.44 = 0.3 + x - (0.3 \cdot x)$.
$0.44 - 0.3 = x - 0.3x$.
$0.14 = 0.7x$.
$x = \frac{0.14}{0.7} = 0.2$.

(B) Let $E_1, E_2, E_3$ be the events of choosing bags $A, B, C$ respectively. Since the bag is chosen at random,$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$.
Let $B$ be the event of drawing a black ball.
The probability of drawing a black ball from bag $A$ is $P(B|E_1) = \frac{3}{2+3} = \frac{3}{5}$.
The probability of drawing a black ball from bag $B$ is $P(B|E_2) = \frac{2}{4+2} = \frac{2}{6} = \frac{1}{3}$.
The probability of drawing a black ball from bag $C$ is $P(B|E_3) = \frac{2}{3+2} = \frac{2}{5}$.
By the law of total probability,$P(B) = P(E_1)P(B|E_1) + P(E_2)P(B|E_2) + P(E_3)P(B|E_3)$.
$P(B) = \frac{1}{3} \times \frac{3}{5} + \frac{1}{3} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{5}$.
$P(B) = \frac{1}{5} + \frac{1}{9} + \frac{2}{15} = \frac{9 + 5 + 6}{45} = \frac{20}{45} = \frac{4}{9}$.

(C) Let $B_1$ be the first basket and $B_2$ be the second basket.
In $B_1$,total fruits = $5 + 7 = 12$.
Probability of picking an apple from $B_1$,$P(A_1) = \frac{5}{12}$.
Probability of picking an orange from $B_1$,$P(O_1) = \frac{7}{12}$.
In $B_2$,total fruits = $4 + 8 = 12$.
Probability of picking an apple from $B_2$,$P(A_2) = \frac{4}{12} = \frac{1}{3}$.
Probability of picking an orange from $B_2$,$P(O_2) = \frac{8}{12} = \frac{2}{3}$.
We want one apple and one orange. This can happen in two mutually exclusive ways:
$1$. Apple from $B_1$ and Orange from $B_2$: $P(A_1) \times P(O_2) = \frac{5}{12} \times \frac{2}{3} = \frac{10}{36}$.
$2$. Orange from $B_1$ and Apple from $B_2$: $P(O_1) \times P(A_2) = \frac{7}{12} \times \frac{1}{3} = \frac{7}{36}$.
Total probability = $\frac{10}{36} + \frac{7}{36} = \frac{17}{36}$.

(D) The probability that event $A_i$ does not occur is given by $P(\bar{A}_i) = 1 - P(A_i)$.
Given $P(A_i) = \frac{1}{1+i}$,we have $P(\bar{A}_i) = 1 - \frac{1}{1+i} = \frac{1+i-1}{1+i} = \frac{i}{1+i}$.
Since $A_i$ are independent events,the probability that none of $A_i$ occurs is the product of the probabilities of their complements:
$P(\text{none of } A_i \text{ occurs}) = P(\bar{A}_1 \cap \bar{A}_2 \cap \ldots \cap \bar{A}_n) = P(\bar{A}_1) \cdot P(\bar{A}_2) \cdot \ldots \cdot P(\bar{A}_n)$.
Substituting the values:
$= \left(\frac{1}{2}\right) \cdot \left(\frac{2}{3}\right) \cdot \left(\frac{3}{4}\right) \cdot \ldots \cdot \left(\frac{n}{n+1}\right)$.
This is a telescoping product where the numerator of each term cancels with the denominator of the previous term:
$= \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \ldots \cdot \frac{n}{n+1} = \frac{1}{n+1}$.

(D) red ball can be drawn in two mutually exclusive ways.
$(i)$ Selecting bag $I$ and then drawing a red ball from it.
(ii) Selecting bag $II$ and then drawing a red ball from it.
Let $E_1$,$E_2$ and $A$ denote the events defined as follows:
$E_1 = \text{Selecting bag } I$
$E_2 = \text{Selecting bag } II$
Since one of the two bags is selected randomly,we have:
$P(E_1) = \frac{1}{2}$ and $P(E_2) = \frac{1}{2}$
Now,$P(A|E_1) = \text{Probability of drawing a red ball when the first bag is selected} = \frac{4}{7}$
$P(A|E_2) = \text{Probability of drawing a red ball when the second bag is selected} = \frac{2}{5}$
Using the law of total probability:
$P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2)$
$P(A) = \frac{1}{2} \times \frac{4}{7} + \frac{1}{2} \times \frac{2}{5}$
$P(A) = \frac{2}{7} + \frac{1}{5} = \frac{10 + 7}{35} = \frac{17}{35}$

(A) Let $U_1$ be the event of selecting the first urn and $U_2$ be the event of selecting the second urn. Since the urns are selected at random,$P(U_1) = P(U_2) = \frac{1}{2}$.
Urn $1$ contains $3$ green and $2$ black balls,so the total number of balls is $5$. The probability of drawing a black ball from Urn $1$ is $P(B|U_1) = \frac{2}{5}$.
Urn $2$ contains $2$ green and $5$ black balls,so the total number of balls is $7$. The probability of drawing a black ball from Urn $2$ is $P(B|U_2) = \frac{5}{7}$.
Using the law of total probability,the probability of drawing a black ball $P(B)$ is given by:
$P(B) = P(U_1) \times P(B|U_1) + P(U_2) \times P(B|U_2)$
$P(B) = \frac{1}{2} \times \frac{2}{5} + \frac{1}{2} \times \frac{5}{7}$
$P(B) = \frac{1}{5} + \frac{5}{14}$
$P(B) = \frac{14 + 25}{70} = \frac{39}{70}$

(C) Let $E_1$ be the event of selecting bag $X$ and $E_2$ be the event of selecting bag $Y$. Since one bag is selected at random,$P(E_1) = P(E_2) = \frac{1}{2}$.
Let $W$ be the event of drawing a white ball.
The probability of drawing a white ball from bag $X$ is $P(W|E_1) = \frac{2}{2+3} = \frac{2}{5}$.
The probability of drawing a white ball from bag $Y$ is $P(W|E_2) = \frac{4}{4+2} = \frac{4}{6} = \frac{2}{3}$.
Using the law of total probability,the probability of drawing a white ball is $P(W) = P(E_1) \cdot P(W|E_1) + P(E_2) \cdot P(W|E_2)$.
$P(W) = \left(\frac{1}{2} \cdot \frac{2}{5}\right) + \left(\frac{1}{2} \cdot \frac{2}{3}\right) = \frac{1}{5} + \frac{1}{3} = \frac{3+5}{15} = \frac{8}{15}$.

(A) bag $B$ contains $4$ white balls and $2$ black balls. Total balls in bag $B = 4 + 2 = 6$.
Probability of selecting $1$ white ball from bag $B$ is $P_1 = \frac{4}{6} = \frac{2}{3}$.
Another bag $C$ contains $3$ white balls and $5$ black balls. Total balls in bag $C = 3 + 5 = 8$.
Probability of selecting $1$ white ball from bag $C$ is $P_2 = \frac{3}{8}$.
Since the events are independent,the probability that both balls drawn are white is $P = P_1 \times P_2 = \frac{2}{3} \times \frac{3}{8} = \frac{6}{24} = \frac{1}{4}$.

(D) Let $B$ be the event that the student is a boy and $G$ be the event that the student is a girl. Let $T$ be the event that the student is taller than $1.6 \ m$.
Given: $P(G) = 0.60$,so $P(B) = 1 - 0.60 = 0.40$.
Probability of a boy being taller than $1.6 \ m$: $P(T|B) = 5 \% = 0.05 = \frac{5}{100}$.
Probability of a girl being taller than $1.6 \ m$: $P(T|G) = 2 \% = 0.02 = \frac{2}{100}$.
Using the Law of Total Probability:
$P(T) = P(B) \cdot P(T|B) + P(G) \cdot P(T|G)$
$P(T) = 0.40 \cdot 0.05 + 0.60 \cdot 0.02$
$P(T) = 0.020 + 0.012 = 0.032$
$P(T) = \frac{32}{1000} = \frac{4}{125}$.

(C) Let $E_1, E_2, E_3$ be the events of selecting bag $A$,bag $B$,and bag $C$ respectively. Since the bag is chosen at random,$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$.
Let $B$ be the event of drawing a black ball.
The probability of drawing a black ball from bag $A$ is $P(B|E_1) = \frac{2}{6} = \frac{1}{3}$.
The probability of drawing a black ball from bag $B$ is $P(B|E_2) = \frac{3}{6} = \frac{1}{2}$.
The probability of drawing a black ball from bag $C$ is $P(B|E_3) = \frac{4}{6} = \frac{2}{3}$.
Using the law of total probability:
$P(B) = P(E_1)P(B|E_1) + P(E_2)P(B|E_2) + P(E_3)P(B|E_3)$
$P(B) = \left(\frac{1}{3} \times \frac{1}{3}\right) + \left(\frac{1}{3} \times \frac{1}{2}\right) + \left(\frac{1}{3} \times \frac{2}{3}\right)$
$P(B) = \frac{1}{9} + \frac{1}{6} + \frac{2}{9} = \frac{3}{9} + \frac{1}{6} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$.

(B) The total number of cards in a deck is $52$,and the number of aces is $4$.
When the first card is drawn,the probability of getting an ace is $P(A_1) = \frac{4}{52} = \frac{1}{13}$.
Since the card is drawn without replacement,there are now $51$ cards left in the deck,and $3$ of them are aces.
The probability of drawing a second ace,given that the first card was an ace,is $P(A_2|A_1) = \frac{3}{51} = \frac{1}{17}$.
The probability that both cards are aces is $P(A_1 \cap A_2) = P(A_1) \times P(A_2|A_1)$.
$P(A_1 \cap A_2) = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$.

(B) Let $U_A, U_B,$ and $U_C$ be the events of choosing urn $A, B,$ and $C$ respectively. Since the urn is chosen at random,$P(U_A) = P(U_B) = P(U_C) = \frac{1}{3}$.
Let $W$ be the event of drawing a white ball.
The conditional probabilities are:
$P(W|U_A) = \frac{6}{6+2} = \frac{6}{8} = \frac{3}{4}$
$P(W|U_B) = \frac{5}{5+3} = \frac{5}{8}$
$P(W|U_C) = \frac{4}{4+4} = \frac{4}{8} = \frac{1}{2}$
Using the Law of Total Probability:
$P(W) = P(U_A)P(W|U_A) + P(U_B)P(W|U_B) + P(U_C)P(W|U_C)$
$P(W) = \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{5}{8} + \frac{1}{3} \times \frac{1}{2}$
$P(W) = \frac{1}{3} \times (\frac{6}{8} + \frac{5}{8} + \frac{4}{8}) = \frac{1}{3} \times \frac{15}{8} = \frac{5}{8}$.

(C) The total number of cards is $52$. There are $4$ kings and $4$ aces in a deck.
Step $1$: Probability of drawing the first king = $4/52 = 1/13$.
Step $2$: Probability of drawing the second king (without replacement) = $3/51 = 1/17$.
Step $3$: Probability of drawing the third card as an ace (without replacement) = $4/50 = 2/25$.
Total probability = $(4/52) \times (3/51) \times (4/50) = (1/13) \times (1/17) \times (2/25) = 2 / 5525$.