Complementary Events and Simple Problems Questions in English

(C) Let $S$ and $R$ denote the events that Sangeeta wins the match and Reshma wins the match,respectively.
The probability of Sangeeta winning is $P(S) = 0.62$ (given).
Since the events $S$ and $R$ are complementary (assuming one must win),the probability of Reshma winning is $P(R) = 1 - P(S)$.
Therefore,$P(R) = 1 - 0.62 = 0.38$.

(A) Out of the two friends,one girl,say,Savita's birthday can be any day of the year. Now,Hamida's birthday can also be any day of $365$ days in the year.
We assume that these $365$ outcomes are equally likely.
$(i)$ If Hamida's birthday is different from Savita's,the number of favourable outcomes for her birthday is $365 - 1 = 364$.
So,$P(\text{Hamida's birthday is different from Savita's birthday}) = \frac{364}{365}$.
$(ii)$ $P(\text{Savita and Hamida have the same birthday})$
$= 1 - P(\text{both have different birthdays})$
$= 1 - \frac{364}{365} \quad [\text{Using } P(\overline{E}) = 1 - P(E)]$
$= \frac{1}{365}$.

(A) $(i)$ The sum of the probabilities of an event $E$ and its complement 'not $E$' is always $1$. Thus,$P(E) + P(\text{not } E) = 1$.
$(ii)$ The probability of an event that cannot happen is $0$. Such an event is called an impossible event.
$(iii)$ The probability of an event that is certain to happen is $1$. Such an event is called a sure event or a certain event.

(A) We know that the sum of the probability of an event and the probability of its complement is $1$.
$P(E) + P(\text{not } E) = 1$
Given that $P(E) = 0.05$.
Therefore,$P(\text{not } E) = 1 - P(E)$
$P(\text{not } E) = 1 - 0.05$
$P(\text{not } E) = 0.95$
Thus,the probability of 'not $E$' is $0.95$.

(C) Let $E$ be the event that $2$ students have the same birthday.
Let $\bar{E}$ be the event that $2$ students do not have the same birthday.
It is given that $P(\bar{E}) = 0.992$.
We know that the sum of the probabilities of complementary events is $1$,so $P(E) + P(\bar{E}) = 1$.
Therefore,$P(E) = 1 - P(\bar{E})$.
$P(E) = 1 - 0.992 = 0.008$.

(C) The possible outcomes when tossing a coin $3$ times are:
$(HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)$
Total number of possible outcomes $= 8$.
Hanif wins if all tosses give the same result,which are $(HHH)$ or $(TTT)$.
Number of favourable outcomes for winning $= 2$.
Probability of winning $P(\text{Win}) = \frac{2}{8} = \frac{1}{4}$.
Since the sum of the probability of an event and its complement is $1$,the probability of losing is:
$P(\text{Lose}) = 1 - P(\text{Win}) = 1 - \frac{1}{4} = \frac{3}{4}$.

(A) Total number of outcomes $= 6 \times 6 = 36$.
$(i)$ Let $E$ be the event that $5$ comes up at least once. The outcomes where $5$ appears are $(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)$.
Number of favourable outcomes for $E = 11$.
Probability of $5$ coming up at least once,$P(E) = \frac{11}{36}$.
The probability that $5$ will not come up either time is $P(\text{not } E) = 1 - P(E) = 1 - \frac{11}{36} = \frac{25}{36}$.
$(ii)$ The probability that $5$ will come up at least once is $P(E) = \frac{11}{36}$.

(A) The sum of the probability of an event and the probability of its complementary event is always $1$.
Let $E$ be an event and $E'$ be its complementary event.
Then,$P(E) + P(E') = 1$.
Given that the probability of the event is $p$,we have $P(E) = p$.
Substituting this into the equation,we get $p + P(E') = 1$.
Therefore,the probability of its complementary event is $P(E') = 1 - p$.

(C) standard deck of cards contains $52$ cards in total.
There is only $1$ card that is the ace of hearts.
The event $E$ is defined as the card drawn is not an ace of hearts.
To find the number of outcomes favourable to $E$,we subtract the number of ace of hearts from the total number of cards.
Number of favourable outcomes = Total cards - Number of ace of hearts
Number of favourable outcomes = $52 - 1 = 51$.

(B) No,this is not correct.
When a fair die is thrown,the total number of possible outcomes is $6$,which are ${1, 2, 3, 4, 5, 6}$.
The probability of getting $1$ is $P(1) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{6}$.
The probability of getting 'not $1$' is $P(\text{not } 1) = 1 - P(1) = 1 - \frac{1}{6} = \frac{5}{6}$.
Since $\frac{1}{6} \neq \frac{5}{6}$,the student's claim that each probability is $\frac{1}{2}$ is incorrect.

(A) When two dice are thrown at the same time,the total number of possible outcomes is $6 \times 6 = 36$.
$(i)$ Let $E_1$ be the event of getting the same number on both dice.
The favorable outcomes are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$.
Number of favorable outcomes $= 6$.
Probability $P(E_1) = \frac{6}{36} = \frac{1}{6}$.
$(ii)$ Let $E_2$ be the event of getting different numbers on both dice.
This is the complement of event $E_1$.
Number of favorable outcomes $= 36 - 6 = 30$.
Probability $P(E_2) = \frac{30}{36} = \frac{5}{6}$.

(B) In the experiment of selecting a card at random from a well-shuffled pack of $52$ cards,the total number of elementary outcomes is $52$.
Let event $D$ be the event that the selected card is not a king.
Let event $\bar{D}$ be the event that the selected card is a king.
The number of outcomes favourable to event $\bar{D}$ is $4$,as there are $4$ kings in a standard pack of cards.
Therefore,$P(\bar{D}) = \frac{4}{52} = \frac{1}{13}$.
Using the complement rule,$P(D) = 1 - P(\bar{D})$.
Therefore,$P(D) = 1 - \frac{1}{13} = \frac{12}{13}$.
Thus,the probability that the selected card is not a king is $\frac{12}{13}$.

(A) Total number of balls in the box $= 5 + 8 + 4 = 17$.
Therefore,the total number of elementary outcomes in the experiment of drawing a ball from the box $= 17$.
Let event $C$ be the event that the ball drawn is not green.
Let event $\overline{C}$ be the event that the ball drawn is green.
There are $4$ green balls in the box.
Therefore,the number of outcomes favourable to event $\overline{C} = 4$.
Thus,$P(\overline{C}) = \frac{4}{17}$.
Using the property of complementary events,$P(C) = 1 - P(\overline{C})$.
Therefore,$P(C) = 1 - \frac{4}{17} = \frac{17 - 4}{17} = \frac{13}{17}$.

(D) Total number of balls = $5 + 7 + 4 + 2 = 18$.
Number of white balls = $5$.
Number of balls that are not white = Total balls - White balls = $18 - 5 = 13$.
Probability of selecting a ball that is not white = $\frac{\text{Number of non-white balls}}{\text{Total number of balls}} = \frac{13}{18}$.

(B) Total number of balls = $5 + 7 + 4 + 2 = 18$.
Number of blue balls = $2$.
Probability of selecting a blue ball $P(\text{Blue}) = \frac{\text{Number of blue balls}}{\text{Total number of balls}} = \frac{2}{18} = \frac{1}{9}$.
The probability that the ball is not blue is given by $P(\text{Not Blue}) = 1 - P(\text{Blue})$.
$P(\text{Not Blue}) = 1 - \frac{1}{9} = \frac{8}{9}$.

(D) The sum of the probability of an event and its complement is always $1$.
Given $P(A) = 0.54$.
We know that $P(A) + P(\overline{A}) = 1$.
Therefore,$P(\overline{A}) = 1 - P(A)$.
$P(\overline{A}) = 1 - 0.54 = 0.46$.

(A) We know that for any event $B$,the sum of the probability of the event and its complement is $1$.
$P(B) + P(\overline{B}) = 1$
Given that $P(\overline{B}) = 0.81$.
Substituting the value,we get:
$P(B) + 0.81 = 1$
$P(B) = 1 - 0.81$
$P(B) = 0.19$

(B) The sum of the probability of an event and its complement is always $1$.
Given $P(C) = \frac{2}{5}$.
We know that $P(C) + P(\bar{C}) = 1$.
Therefore,$P(\bar{C}) = 1 - P(C)$.
$P(\bar{C}) = 1 - \frac{2}{5} = \frac{5 - 2}{5} = \frac{3}{5}$.

(C) The sum of the probability of an event and its complement is always $1$.
That is,$P(D) + P(\overline{D}) = 1$.
Given that $P(\overline{D}) = \frac{6}{17}$,we can substitute this value into the equation:
$P(D) + \frac{6}{17} = 1$
$P(D) = 1 - \frac{6}{17}$
$P(D) = \frac{17 - 6}{17}$
$P(D) = \frac{11}{17}$

(B) We know that the sum of the probability of an event and its complement is $1$,i.e.,$P(A) + P(\bar{A}) = 1$.
Given the ratio $P(A) : P(\bar{A}) = 3 : 2$,let $P(A) = 3x$ and $P(\bar{A}) = 2x$.
Substituting these into the equation: $3x + 2x = 1$.
$5x = 1$,which gives $x = \frac{1}{5}$.
Therefore,$P(A) = 3x = 3 \times \frac{1}{5} = \frac{3}{5}$.

(C) We know that for any event $A$,the sum of the probability of the event and its complement is $1$,i.e.,$P(A) + P(\overline{A}) = 1$.
Given the ratio $P(A) : P(\overline{A}) = 4 : 1$,we can let $P(A) = 4x$ and $P(\overline{A}) = 1x$ for some constant $x$.
Substituting these into the sum equation: $4x + 1x = 1$.
$5x = 1$,which gives $x = \frac{1}{5}$.
Therefore,$P(\overline{A}) = 1x = 1 \times \frac{1}{5} = \frac{1}{5}$.

(C) In a standard pack of $52$ cards,there are $4$ suits,each containing $13$ cards.
The number of Diamond cards is $13$.
The number of cards that are not Diamonds is $52 - 13 = 39$.
The probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
$\text{Probability (not a Diamond)} = \frac{\text{Number of non-Diamond cards}}{\text{Total number of cards}} = \frac{39}{52}$.
Simplifying the fraction by dividing both numerator and denominator by $13$,we get $\frac{39 \div 13}{52 \div 13} = \frac{3}{4}$.

(A) In a standard pack of $52$ cards,there are $4$ aces.
Therefore,the number of cards that are not aces is $52 - 4 = 48$.
The probability of drawing a card that is not an ace is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
$\text{Probability} = \frac{\text{Number of non-ace cards}}{\text{Total number of cards}} = \frac{48}{52}$.
Simplifying the fraction by dividing both numerator and denominator by $4$,we get $\frac{48 \div 4}{52 \div 4} = \frac{12}{13}$.

(B) Total number of balls in the bag $= 6 + 5 + 4 = 15$.
$\therefore$ Number of total outcomes $= n = 15$.
The number of outcomes favouring the event that the ball drawn is not red $= m = 6 \text{ (green)} + 4 \text{ (blue)} = 10$.
$\therefore$ Required probability $= \frac{m}{n} = \frac{10}{15} = \frac{2}{3}$.

(C) The sum of the probabilities of an event and its complement is always $1$.
Given that $P(A) = 0.6$.
We know that $P(A) + P(\bar{A}) = 1$.
Therefore,$P(\bar{A}) = 1 - P(A)$.
$P(\bar{A}) = 1 - 0.6 = 0.4$.

(C) The sum of the probability of an event and its complement is always $1$.
Given that $P(\bar{B}) = 0.55$.
We know that $P(B) + P(\bar{B}) = 1$.
Therefore,$P(B) = 1 - P(\bar{B})$.
Substituting the value: $P(B) = 1 - 0.55 = 0.45$.
Thus,the correct option is $C$.

(B) Given the ratio $P(A) : P(\bar{A}) = 3 : 4$.
Let $P(A) = 3x$ and $P(\bar{A}) = 4x$,where $x$ is a constant.
We know that the sum of the probability of an event and its complement is always $1$,i.e.,$P(A) + P(\bar{A}) = 1$.
Substituting the values,we get $3x + 4x = 1$.
$7x = 1$,which implies $x = \frac{1}{7}$.
Therefore,$P(A) = 3x = 3 \times \frac{1}{7} = \frac{3}{7}$.

(D) We know that for any event $A$,the sum of the probability of the event and its complement is $P(A) + P(\bar{A}) = 1$.
Given the ratio $P(A) : P(\bar{A}) = 5 : 3$,we can assume $P(A) = 5k$ and $P(\bar{A}) = 3k$ for some constant $k$.
Substituting these into the identity: $5k + 3k = 1$.
$8k = 1$,which implies $k = \frac{1}{8}$.
Therefore,$P(\bar{A}) = 3k = 3 \times \frac{1}{8} = \frac{3}{8}$.

(B) Let event $A$ be the event that the numbers appearing on both the dice are different.
Then,the complementary event $\overline{A}$ is the event that the numbers appearing on both the dice are the same.
The total number of possible outcomes when two dice are rolled is $6 \times 6 = 36$.
The outcomes favorable to event $\overline{A}$ are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$.
Thus,the number of favorable outcomes for $\overline{A}$ is $m = 6$.
The probability of event $\overline{A}$ is $P(\overline{A}) = \frac{m}{n} = \frac{6}{36} = \frac{1}{6}$.
Since $P(A) + P(\overline{A}) = 1$,the probability that the numbers are different is $P(A) = 1 - P(\overline{A}) = 1 - \frac{1}{6} = \frac{5}{6}$.

(D) We know that the sum of the probability of an event and its complement is $1$,i.e.,$P(A) + P(\bar{A}) = 1$.
We are given the equation: $P(A) - P(\bar{A}) = 0.5$.
Adding these two equations:
$(P(A) + P(\bar{A})) + (P(A) - P(\bar{A})) = 1 + 0.5$
$2P(A) = 1.5$
Dividing by $2$:
$P(A) = \frac{1.5}{2} = 0.75$.