Probability

Galileo an Italian mathematician was the first person to attempt at a quantitative measure of probability while dealing with some problems related to the theory of dice in gambling. Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty we can predict only the total chance of an event to occur i.e., how likely they are going to happen, using it.

Ex. (i) It will probably rain today. (ii) India's cricket team has good chances of winning the world-cup.

1.0Some basic concepts/terms

Experiment: An action or operation which can produce some well-defined result is known as experiment.

Deterministic experiment: If we perform an experiment and repeat it under identical conditions, we get almost the same result every time, such an experiment is called a deterministic experiment.

Random experiment: An experiment is said to be a random experiment if it satisfies the following two conditions: (i) It has more than one possible outcome. (ii) It is not possible to predict the outcome (result) in advance. E.g., (i) Tossing a pair of fair coins. (ii) Rolling an unbiased die.

Outcomes: The possible results of a random experiment are called outcomes. Trial: When an experiment is repeated under similar conditions, and it does not give the same result each time but may result in any one of the several possible outcomes is called a trial. E.g., If a coin is tossed 100 times, then one toss of the coin is called a trial.

Event: The collection of all or some outcomes of a random experiment is called an event. E.g., Suppose we toss a pair of coins simultaneously and let E be the event of getting exactly one head. Then, the event E contains HT and TH. E.g., Suppose we roll a die and let E be the event of getting an even number. Then the event E contains 2,4 and 6 .

• When we toss ' $n$ ' coins simultaneously or when we toss a coin ' $n$ ' times then number of possible outcomes are $2^n$. Elementary or Simple Event: An outcome of a trial is called an elementary event. Note: An elementary event has only one element. E.g., Let a pair of coins is tossed simultaneously.

Then, possible outcomes of this experiment are HH : Getting H on first and H on second $\left(={\mathrm{E}}_1\right)$

$\text{ [ }\mathrm{H}=\text{ Head, }\mathrm{T}=\text{ Tail and }\mathrm{E}=\text{ event }]$

HT : Getting H on first and $T$ on second $\left(=E_2\right)$ TH : Getting $T$ on first and $H$ on second $\left(=E_3\right)$ and TT : Getting $T$ on first and $T$ on second $\left(=E_4\right)$ Here, ${\mathrm{E}}_1,{\mathrm{E}}_2,{\mathrm{E}}_3$ and ${\mathrm{E}}_4$ are the elementary events associated with the random experiment of tossing of two coins.

Compound event or composite event or mixed event

An event associated to a random experiment and obtained by combining two or more simple events associated to the same random experiment, is called a compound event.

A compound event is an aggregate of some simple (elementary) event and is decomposable into simple events. e.g., If we throw a die, then the event $E$ of getting an odd number is a compound event because the event $E$ contains three elements 1,3 and 5 , which is a compound of three simple events $E_1,E_2$ and ${\mathrm{E}}_3$ containing 1, 3 and 5 respectively.

Equally likely events: The outcomes of an experiment are said to be equally likely events if the chances of their happenings are neither less nor greater than other.

In other words, a given number of events are said to be equally likely if none of them is expected to occur in preference to the others. E.g., In tossing a coin, getting head (H) and tail (T) are equally likely events.

Building Concepts 1 Which of the following experiments have equally likely outcomes? Explain. (i) A coin is tossed. It turns to be a head or a tail. (ii) A monitor is nominated by the class teacher of a class. It is a boy or a girl. Explanation: (i) Yes, it is equally likely outcome because when a coin is tossed either a head or a tail turns. (ii) Yes, it is an equally likely outcome because when a class teacher nominates for the monitor, the student will either be a boy or a girl.

• When we throw ' n ' dice simultaneously or when we throw a dice ' n ' times, then number of possible outcomes are $6^n$.

2.0Experimental (or empirical) probability

The experimental or empirical probability $\mathrm{P}(\mathrm{E})$ of an event is defined as $P(E)=\frac{\text{ Number of trials in which the event happened }}{\text{ Total numbers of trials }}$ i.e., $P(E)=\frac{m}{n}$ Clearly, $0\le m\le n$ $0\le \frac{\mathrm{m}}{\mathrm{n}}\le 1$ $\Rightarrow 0\le \mathrm{P}(\mathrm{E})\le 1$ Probability of an event always lies from 0 to 1 .

• These probabilities are based on results of an actual experiment.
• These probabilities are only 'estimates', i.e., we may get different probabilities for the same event in various experiments.

3.0Theoretical (or classical) probability

The theoretical or classical probability of an event E, written as $\mathrm{P}(\mathrm{E})$, is defined as $P(E)=\frac{\text{ Number of outcomes favourable to E }}{\text{ Number of all possible outcomes of the experiment }}$ where the outcomes of the experiment are equally likely.

Building Concepts 2 A die is thrown once (i) What is the probability of getting a number greater than 4 ? (ii) What is the probability of getting a number less than or equal to 4? Explanation: The possible outcomes are $1,2,3,4,5$ and 6 . Let $\mathrm{E}=$ The event of getting a number greater than 4 and $\mathrm{F}=$ The event of getting a number less than or equal to 4 . (i) The outcomes favourable to E are 5 and 6. $\therefore$ The number of outcomes favourable to E is 2 . Therefore, $P(E)=P($ number greater than 4$)=\frac{2}{6}=\frac{1}{3}$ (ii) The outcomes favourable to F are 1, 2, 3, 4. $\therefore$ The number of outcomes favourable to F is 4 . Therefore, $\mathrm{P}(\mathrm{E})=\mathrm{P}($ number less than or equal to 4$)=\frac{4}{6}=\frac{2}{3}$

Note : Events E and F are not elementary events because event E has 2 outcomes and the event F has 4 outcomes.

Building Concepts 3 Two unbiased coins are tossed simultaneously. Find the probability of getting (i) one head (ii) one tail (iii) two heads (iv) at least one head (v) at most one head (vi) no head Explanation: If two unbiased coins are tossed simultaneously, then all possible outcomes are HH, HT, TH, TT. Total number of possible outcomes $=4$. (i) Let ${\mathrm{A}}_1=$ The event of getting one head. Then, favourable outcomes are HT, TH. Number of favourable outcomes $=2$. Hence, required probability $=\mathrm{P}($ getting one head $)=P\left({\mathrm{A}}_1\right)=\frac{2}{4}=\frac{1}{2}$ (ii) Let ${\mathrm{A}}_2=$ The event of getting one tail. Then, favourable outcomes are TH, HT. Number of favourable outcomes $=2$. Hence, required probability $=\mathrm{P}($ getting one tail $)=\mathrm{P}\left({\mathrm{A}}_2\right)=\frac{2}{4}=\frac{1}{2}$ (iii) Let ${\mathrm{A}}_3=$ The event of getting two heads Then, the favourable outcome is HH Number of favourable outcome $=1$ Hence, required probability $=\mathrm{P}($ getting two heads $)=P\left({\mathrm{A}}_3\right)=\frac{1}{4}$ (iv) Let ${\mathrm{A}}_4=$ The event of getting at least one head. Then, the favourable outcomes are HT, TH, HH Number of favourable outcomes = 3 Hence, required probability $=\mathrm{P}($ getting at least one head $)=P\left({\mathrm{A}}_4\right)=\frac{3}{4}$ (v) Let ${\mathrm{A}}_5=$ The event of getting atmost one head. Then, the favourable outcomes are TT, HT, TH. Number of favourable outcomes $=3$ Hence, required probability $=P($ getting atmost one head $)=P\left(A_5\right)=\frac{3}{4}$ (vi) Let ${\mathrm{A}}_6=$ The event of getting no head. Then, the favourable outcomes is TT Number of favourable outcome $=1$ Hence, required probability $=P($ getting no head $)=P\left(A_6\right)=\frac{1}{4}$

4.0Some special events

Impossible event (or null event)

An event is said to be an impossible event when none of the outcomes is favourable to the event. The probability of an impossible event $=0$.

Building Concepts 4 What is the probability of getting a number 8 in a single throw of a die? Explanation: The possible outcomes are $1,2,3,4,5,6$. Let $\mathrm{E}=$ The event of getting a number 8 in a single throw of a die. Clearly, the number of outcomes favourable to E is 0 and the total number of possible outcomes is 6 . Therefore, $P(E)=\frac{0}{6}=0$. Here, $E$ is an impossible event.

Sure (or certain) event

An event is said to be a sure (or certain) event when all possible outcomes are favourable to the event. The probability of a sure event is 1 .

Building Concepts 5 What is the probability of getting a number less than 7 in a single throw of a die? Explanation: The possible outcomes are: $1,2,3,4,5,6$. Let $F=$ the event of getting a number less than 7 in a single throw of a die. Clearly, the number of outcomes favourable to F are $1,2,3,4,5,6$. i.e., the number of outcomes favourable to F is 6 . Therefore, $P(F)=\frac{6}{6}=1$. Here, $F$ is a sure (or certain) event.

Complement of an event

Corresponding to every event E associated with random experiment, there is an event 'not E ', which occurs only when E does not occur. The event $\overline{\mathrm{E}}$, representing 'not E ', is called the complement of the event E . $E$ and $\bar{E}$ are also called complementary events. In general, $\mathbf{P}(\mathbf{E})+\mathbf{P}(\overline{\mathrm{E}})=1$ i.e., $P(\bar{E})=1-P(E)$ or $P(\mathrm{not}E)=1-P(E)$

• Let $E_1,E_2,E_3,\dots ,E_n$ be the $n$ elementary events associated with a random experiment having exactly $n$ outcomes. Then, $\mathrm{P}\left({\mathrm{E}}_1\right)+\mathrm{P}\left({\mathrm{E}}_2\right)+\mathrm{P}\left({\mathrm{E}}_3\right)+\dots +\mathrm{P}\left({\mathrm{E}}_{\mathrm{n}}\right)=1$ E.g., A bag contains 3 red balls, 4 white balls and 5 green balls. A ball is drawn at random.

Let $\mathrm{R}=$ The event of getting a red ball, $\mathrm{W}=$ The event of getting a white ball and $\mathrm{G}=$ The event of getting a green ball. Here, total number of balls (outcomes) $=3+4+5=12$. Then, $P(R)=\frac{3}{12}$ [Number of favourable outcomes $=3$ ] $P(W)=\frac{4}{12}$ [Number of favourable outcomes $=4]$ and $\mathrm{P}(\mathrm{G})=\frac{5}{12}$ [Number of favourable outcomes $=5$ ] Clearly, $\frac{3}{12}+\frac{4}{12}+\frac{5}{12}=\frac{12}{12}=1$ i.e., $P(R)+P(W)+P(G)=1$.

Numerical Ability 1 If $P(E)=0.05$, what is the probability of 'not $E$ '? Solution: We have $\mathrm{P}(\mathrm{E})=0.05$ $\therefore P($ not $E)=1-P(E)=1-0.05=0.95$ Therefore, $\mathrm{P}(\mathrm{not}\mathrm{E})=0.95$.

Designation of playing cards

Building Concepts 6 A card is drawn at random from a well shuffled deck of 52 cards. Find the probability that the card drawn is (i) a red card (ii) a non-ace (iii) a king or a jack (iv) neither a king nor a queen Explanation: If a card is drawn at random from a well shuffled deck of 52 cards, then total number of possible outcomes $=52$ (i) Let ${\mathrm{A}}_1=$ event of getting a red card. Then, the number of favourable outcomes $=26$. Hence, required probability $=P$ (getting a red card)$=P\left(A_1\right)=\frac{26}{52}=\frac{1}{2}$ (ii) Let ${\mathrm{A}}_2=$ Event of getting a non-ace. Then, the number of favourable outcomes $=48$. [ $because$thereare4acesinapackofplayingcards]Hence,requiredprobability$=P$(gettinganon-ace)$$=\frac{48}{52}=\frac{12}{13}$$(iii)Let$A_{3}=$EventofgettingakingorajackThereare4kingcardsand4jackcards.Hence,requiredprobability$=\mathrm{P}\left(\mathrm{A}{3}\right)=\mathrm{P}(getting a king or a jack)=P(getting a king)+P(getting a jack)=\frac{4}{52}+\frac{4}{52}=\frac{8}{52}=\frac{2}{13}(iv) Let\mathrm{A}{4}=$Eventofgettingneitherakingnoraqueen.Thereare4kingand4queencards.Hence,requiredprobability$=\mathrm{P}\left(\mathrm{A}4\right)=\mathrm{P}(getting neither a king nor a queen) =1-\mathrm{P}(getting a king or a queen)=1-[\mathrm{P}(getting a king)+\mathrm{P}(getting a queen)] =1-\left(\frac{4}{52}+\frac{4}{52}\right)=1-\frac{8}{52}=\frac{44}{52}=\frac{11}{13}Alternate method : Let\mathrm{A}{4}$:Eventofgettingneitherkingnorqueen.Thereare4kingand4queencards.$\therefore \quad$No.offavourableoutcomes,i.e.,neitherkingnorqueencards$=52-8=44$Hence,$\mathrm{P}\left(\mathrm{A}_{4}\right)=\frac{44}{52}=\frac{11}{13}$

Building Concepts 7 A box contains 5 red, 4 green and 7 white balls. A ball is drawn at random from the box. Find the probability that the ball drawn is (i) White (ii) Neither red nor white Explanation: Total number of balls in the box $=5+4+7=16$. Let ${\mathrm{A}}_1=$ Event of getting a white ball, ${\mathrm{A}}_2=$ Event of getting neither red nor white ball. (i) There are 7 white balls in the box. Hence, required probability $=P\left(A_1\right)=P$ (getting a white ball) $=\frac{7}{16}$ (ii) There are 7 white and 5 red balls in the box. Hence, required probability $=P$ (getting neither red nor white ball) $=P\left(A_2\right)$ $=1-\mathrm{P}$ (getting either red or white ball) $=1-\mathrm{P}$ (getting a red ball) $+\mathrm{P}$ (getting a white ball) $=1-\left(\frac{5}{16}+\frac{7}{16}\right)=1-\frac{12}{16}=\frac{4}{16}=\frac{1}{4}$ Alternate method : $\mathrm{P}$ (getting neither red nor white ball) $=\mathrm{P}$ getting a green ball)$=\frac{4}{16}=\frac{1}{4}$ $P(\mathrm{not}E)=0.93$

5.0Memory map