Probability Class 10: Events, Outcomes and the Basic Formula

Equally likely out­comes#

Out­comes are equally likely when none of them is expected to occur more often than any other. A fair coin, an unbi­ased die and a well-shuf­fled pack of cards all give equally likely out­comes. Words such as "fair", "unbi­ased", "at ran­dom" and "well-shuf­fled" in a ques­tion are telling you that the out­comes are equally likely. The for­mula below works only in this sit­u­a­tion.

Event#

An event is a col­lec­tion of out­comes that we are inter­ested in. When a die is rolled, "get­ting an even num­ber" is an event made up of the out­comes 2, 4 and 6. The out­comes that make an event hap­pen are called the out­comes favourable to it.

An event with only one out­come, such as "get­ting a 5", is called an ele­men­tary event.

The for­mula for the­o­ret­i­cal prob­a­bil­ity#

State­ment#

If an exper­i­ment has $n$ equally likely out­comes and $m$ of them are favourable to an event $E$, then the prob­a­bil­ity of $E$ is

$$\displaystyle P(E) = \frac{\text{number of outcomes favourable to } E}{\text{total number of possible outcomes}} = \frac{m}{n}.$$

For exam­ple, when a die is rolled, three of the six out­comes (2, 4 and 6) are even, so $\displaystyle P(\text{even}) = \frac{3}{6} = \frac{1}{2}$.

The range of a prob­a­bil­ity#

Since $m$ counts some of the $n$ out­comes, we always have $0 \leq m \leq n$. Divid­ing by the pos­i­tive num­ber $n$ gives

$$0 \leq P(E) \leq 1.$$

So a prob­a­bil­ity can never be neg­a­tive and can never be greater than 1. An answer such as $\displaystyle \frac{7}{5}$ or $-0.2$ is a sig­nal that some­thing has gone wrong in the count­ing.

Impos­si­ble and sure events#

An event that has no favourable out­comes is an impos­si­ble event; here $m = 0$, so its prob­a­bil­ity is 0. Get­ting a 7 on an ordi­nary die is impos­si­ble.

An event for which every out­come is favourable is a sure (cer­tain) event; here $m = n$, so its prob­a­bil­ity is 1. Get­ting a num­ber less than 7 on a die is sure.

Com­ple­men­tary events#

The rule#

For any event $E$, the event "not $E$" (writ­ten $\overline{E}$) hap­pens exactly when $E$ does not. The two are called com­ple­men­tary events. The key result is

$$P(E) + P(\overline{E}) = 1, \quad \text{so} \quad P(\overline{E}) = 1 - P(E).$$

Why it is true#

Sup­pose there are $n$ equally likely out­comes and $m$ of them are favourable to $E$. Every out­come is either favourable to $E$ or not, and never both. So the remain­ing $n - m$ out­comes are exactly the ones favourable to $\overline{E}$. Hence

$$\displaystyle P(E) + P(\overline{E}) = \frac{m}{n} + \frac{n - m}{n} = \frac{n}{n} = 1.$$

Ele­men­tary events add up to 1#

The same argu­ment shows that the prob­a­bil­i­ties of all the ele­men­tary events of an exper­i­ment add up to 1. For a die, each of the six ele­men­tary events has prob­a­bil­ity $\displaystyle \frac{1}{6}$, and

$$\displaystyle \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = 1.$$

This is a use­ful check: if the prob­a­bil­i­ties of all the sep­a­rate out­comes do not add to 1, a mis­take has been made.

Method: find­ing a prob­a­bil­ity#

1. Read the ques­tion and check that the out­comes are equally likely (fair, unbi­ased, at ran­dom).
2. List or count all the pos­si­ble out­comes to get $n$.
3. Write down the event in words, then list or count the out­comes favourable to it to get $m$.
4. Write $\displaystyle P(E) = \frac{m}{n}$ and sim­plify the frac­tion.
5. Check that the answer lies between 0 and 1. If the ques­tion asks for "not" some­thing, it is often quicker to use $P(\overline{E}) = 1 - P(E)$.

Worked exam­ples#

Exam­ple 1: one die#

A fair die is rolled once. Find the prob­a­bil­ity of get­ting (a) a num­ber greater than 4, (b) a num­ber not greater than 4.

The out­comes are 1, 2, 3, 4, 5, 6, so $n = 6$.

(a) The num­bers greater than 4 are 5 and 6, so $m = 2$ and

$$\displaystyle P(\text{greater than } 4) = \frac{2}{6} = \frac{1}{3}.$$

(b) "Not greater than 4" is the com­ple­ment of the event in (a), so

$$\displaystyle P(\text{not greater than } 4) = 1 - \frac{1}{3} = \frac{2}{3}.$$

Check by count­ing: 1, 2, 3, 4 give $\displaystyle \frac{4}{6} = \frac{2}{3}$.

Answer: (a) $\displaystyle \frac{1}{3}$ (b) $\displaystyle \frac{2}{3}$

Exam­ple 2: a bag of balls#

A bag holds 5 red, 3 blue and 4 green balls of the same size. One ball is taken out at ran­dom. Find the prob­a­bil­ity that it is (a) red, (b) not green, (c) yel­low.

Total num­ber of balls: $n = 5 + 3 + 4 = 12$.

(a) $\displaystyle P(\text{red}) = \frac{5}{12}$.

(b) The balls that are not green are the 5 red and 3 blue ones, so $m = 8$ and $\displaystyle P(\text{not green}) = \frac{8}{12} = \frac{2}{3}$.

(c) There are no yel­low balls, so this is an impos­si­ble event: $P(\text{yellow}) = 0$.

Check: $\displaystyle P(\text{red}) + P(\text{blue}) + P(\text{green}) = \frac{5}{12} + \frac{3}{12} + \frac{4}{12} = 1$.

Answer: (a) $\displaystyle \frac{5}{12}$ (b) $\displaystyle \frac{2}{3}$ (c) $0$

Exam­ple 3: num­bered cards#

Thirty cards are num­bered 1 to 30 and shuf­fled well. One card is drawn. Find the prob­a­bil­ity that the num­ber on it is (a) a mul­ti­ple of 4, (b) a per­fect square, (c) not a prime num­ber.

Here $n = 30$.

(a) Mul­ti­ples of 4: 4, 8, 12, 16, 20, 24, 28. So $m = 7$ and $\displaystyle P = \frac{7}{30}$.

(b) Per­fect squares: 1, 4, 9, 16, 25. So $m = 5$ and $\displaystyle P = \frac{5}{30} = \frac{1}{6}$.

(c) Primes up to 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, which is 10 num­bers. So

$$\displaystyle P(\text{prime}) = \frac{10}{30} = \frac{1}{3}, \qquad P(\text{not prime}) = 1 - \frac{1}{3} = \frac{2}{3}.$$

Note that 1 is nei­ther prime nor com­pos­ite, so it cor­rectly falls in "not prime".

Answer: (a) $\displaystyle \frac{7}{30}$ (b) $\displaystyle \frac{1}{6}$ (c) $\displaystyle \frac{2}{3}$

Exam­ple 4: using the com­ple­ment with dec­i­mals#

The prob­a­bil­ity that a bulb picked at ran­dom from a batch is defec­tive is 0.08. What is the prob­a­bil­ity that it is not defec­tive? In a match, the prob­a­bil­ity that a team wins is 0.35 and a draw is not pos­si­ble. What is the prob­a­bil­ity that the team loses?

"Defec­tive" and "not defec­tive" are com­ple­men­tary, so

$$P(\text{not defective}) = 1 - 0.08 = 0.92.$$

With no draw, "lose" is the same as "not win", so $P(\text{lose}) = 1 - 0.35 = 0.65$.

Answer: 0.92 and 0.65

Exam­ple 5: a qual­ity check#

A car­ton holds 250 pens, of which 20 are defec­tive. A shop­keeper takes one pen at ran­dom. Find the prob­a­bil­ity that it is (a) good, (b) defec­tive.

Good pens: $250 - 20 = 230$.

$$\displaystyle P(\text{good}) = \frac{230}{250} = \frac{23}{25}, \qquad P(\text{defective}) = \frac{20}{250} = \frac{2}{25}.$$

Check: $\displaystyle \frac{23}{25} + \frac{2}{25} = 1$.

Answer: (a) $\displaystyle \frac{23}{25}$ (b) $\displaystyle \frac{2}{25}$

Exam­ple 6: out­comes that are not equally likely#

Two fair coins are tossed together. Ravi says: "There are three results, two heads, two tails, or one of each, so the prob­a­bil­ity of one head and one tail is $\displaystyle \frac{1}{3}$." Is he right?

No. His three results are not equally likely. List the out­comes of the two coins sep­a­rately: HH, HT, TH, TT. These four are equally likely, so $n = 4$. The event "one head and one tail" is made up of HT and TH, so $m = 2$ and

$$\displaystyle P(\text{one head and one tail}) = \frac{2}{4} = \frac{1}{2}, \qquad P(\text{two heads}) = \frac{1}{4}.$$

The result "one of each" can hap­pen in two ways, which is why it is twice as likely as "two heads".

Answer: Ravi is wrong; the prob­a­bil­ity is $\displaystyle \frac{1}{2}$.

Com­mon mis­takes#

• Using the for­mula when the out­comes are not equally likely, as in Exam­ple 6. Always break the exper­i­ment into out­comes that really are equally likely.
• Giv­ing an answer greater than 1 or less than 0. This always means a count­ing error.
• Count­ing 1 as a prime num­ber. The primes start at 2.
• For­get­ting to sim­plify: $\displaystyle \frac{8}{12}$ should be writ­ten as $\displaystyle \frac{2}{3}$.
• Writ­ing $P(\overline{E}) = P(E) - 1$ instead of $1 - P(E)$.
• Con­fus­ing "at least" and "at most", or "greater than" and "greater than or equal to". Write the favourable out­comes out in full before count­ing.

Try these#

1. A fair die is rolled. Find the prob­a­bil­ity of get­ting an even num­ber less than 5. Answer: $\displaystyle \frac{1}{3}$
2. A let­ter is cho­sen at ran­dom from the word PROB­A­BIL­ITY. Find the prob­a­bil­ity that it is (a) a vowel, (b) the let­ter B. Answer: (a) $\displaystyle \frac{4}{11}$ (b) $\displaystyle \frac{2}{11}$
3. If $P(E) = 0.27$, find $P(\text{not } E)$. Answer: 0.73
4. A box has 6 white and 9 black mar­bles. One is picked at ran­dom. Find the prob­a­bil­ity that it is black. Answer: $\displaystyle \frac{3}{5}$
5. A num­ber is cho­sen at ran­dom from 1 to 50. Find the prob­a­bil­ity that it is a mul­ti­ple of 7. Answer: $\displaystyle \frac{7}{50}$
6. A spin­ner has 8 equal sec­tors num­bered 1 to 8. Find the prob­a­bil­ity that the pointer stops on a mul­ti­ple of 3. Answer: $\displaystyle \frac{1}{4}$

Key terms#

Ran­dom exper­i­ment

An action whose pos­si­ble results are known but whose actual result can­not be pre­dicted.

Out­come

A sin­gle pos­si­ble result of an exper­i­ment.

Equally likely out­comes

Out­comes none of which is expected to occur more often than another.

Event

A col­lec­tion of out­comes of an exper­i­ment.

Ele­men­tary event

An event that con­sists of exactly one out­come.

Impos­si­ble event

An event with no favourable out­comes; its prob­a­bil­ity is 0.

Sure event

An event for which every out­come is favourable; its prob­a­bil­ity is 1.

Com­ple­men­tary events

An event and its "not" event; their prob­a­bil­i­ties add up to 1.

Com­mon ques­tions#

Can a prob­a­bil­ity be 1.5 or neg­a­tive?#

No. The num­ber of favourable out­comes is between 0 and the total num­ber of out­comes, so every prob­a­bil­ity lies between 0 and 1, inclu­sive.

What is the dif­fer­ence between an ele­men­tary event and an event?#

An ele­men­tary event has exactly one out­come, such as "get­ting a 3". An event in gen­eral can have sev­eral out­comes, such as "get­ting an odd num­ber".

Why does the ques­tion say "at ran­dom" or "well-shuf­fled"?#

Those words tell you the out­comes are equally likely. The for­mula favourable over total is valid only under that con­di­tion.

When should I use the com­ple­ment rule?#

Use it when the "not" event is eas­ier to count, or when the ques­tion gives $P(E)$ as a dec­i­mal or frac­tion and asks for $P(\text{not } E)$.

Is the prob­a­bil­ity the same as what will actu­ally hap­pen?#

No. A prob­a­bil­ity of $\displaystyle \frac{1}{2}$ for a head does not mean exactly half of any ten tosses will be heads. It describes what to expect over a very large num­ber of tri­als.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.