# How to Solve Probability Questions? Tips & Tricks

By BYJU'S Exam Prep

Updated on: September 25th, 2023

Hello Friends,

Today we are presenting an article on “How to Solve Probability Questions in competitive examination. **Probability questions** are an important part of **Quantitative aptitude** section of most competitive exams like **SBI, IBPS, PO/Clerk, LIC-AAO etc**. These questions are asked frequently so it becomes really relevant to know the right technique of solving these questions.

**What is Probability?**

‘Probability’ in simple terms tell us about the** chance of something occurring**. The probability of an event happening ranges

**between 0 to 1.**That means the value of probability can never be a negative number or a number greater than 1.

**Consider this,** if it’s cloudy outside then two things can happen. **First, either it will rain or second**, **it won’t rain.** So, the total events are ‘2’ (raining or not raining). **And, the probability of raining is 1/2**

So, **Probability of an event happening = Concerned Events / Total Events**

Probability of an event **happening** is denoted by **P(E)**

Probability of an event **not** **happening** is denoted by **P(Ē)**.

**And**, **P(E) + P(Ē) = 1**

**Types of Events:1. If ‘And’ event is given then we multiply or count events together.2. If ‘Or’ event is given then we ‘add’ the two or more events.**

**Classical Cases:**The classical cases discussed in probability are based on following points:

__1. Dice:__

__1. Dice:__

The dice used here is the one used to play ‘**Ludo**’. A typical dice has numbers **1, 2, 3, 4, 5 and 6 are written over its six faces as shown below.**

When a dice is thrown the** number that appears on upper face is the concerned event.** The questions based on dice are mainly of two types(not exhaustive):

__ (I) When Only One dice is thrown once:__In such cases, the number rolling on playing a

**dice is either 1 or 2 or 3 or 4 or 5 or 6.**

Here,

**the concerned event of rolling out ‘1’ is 1 only because 1 is written only one face.**And, total events = 6 because total different numbers written different faces are six in total.

So,

**Probability of rolling number 1 = 1/6**

Similarly, Probability of rolling number **2 ( or any number from 3 to 6) = 1/6**

**Question asked:****What is the probability of getting an even number on rolling a dice?**

Now, concerned event should have an even number which are 2, 4 and 6

So, Concerned Event = 3

Total Event = 6**So, Probability = 3/6 = 1/2**

__ (II) When Two dices are thrown:__In such cases, either of the two things happen either two dices are thrown simultaneously and the numbers appearing on top faces of both dices are noted and summed up; or one dice is thrown two times in a row and the numbers appearing on the top faces in the two times are noted and summed up. Whatever is done, the treatment is same in either the cases, two dices at once or rolling one dice twice. So,

**this**

__summed up number is the concerned event__in such question.The various combinations of numbers that can turn up on throwing two dices (or one dice twice) can be listed as below –

**For example: (1, 6)** shows that ‘1’ would turn up on dice 1 and ‘6’ would turn up on dice 2. Here, **total outcome** is the total number of combinations stated above: (1,1,); (1,2) ……. (6,5); (6,6) = **36**

**The questions asked can be of following type:****Qs. 1 – ****What is the probability of getting a combination of ‘5’ and ‘3’ on throwing two dices?**

**Solutions – **Now the concerned event should have number ‘5’ and ‘3’ so **concerned events = 2**{(3,5) and (5,3)}

Total events = 36

⇒ **Probability = 2/36 = 1/18**

**Qs. 2 – What is the probability of getting a sum of ‘10’ on rolling a dice twice?**

**Solutions –** Now, the concerned events should **have sum of 10 i.e.** **No. on dice 1 +No. on dice 2 = 10 **

This can be seen in these cases: (4,6); (5,5) and (6,4)

So, Concerned Events = 3

And, we know Total Events are always ‘36’**So, Probability = 3/36 = 1/12**

**2. Coins:**

Coin is a currency token which has two faces, one is head and other is tail. So, when throw a coin in air and when it lands it might have either a head or tail. Coin questions can be three types as shown below:

__I. One Coin once__:

When a coin is tossed is only once then there can be two outcomes either a head or a tail. In such cases, total events = 2**Question: What is the probability of getting a head in a toss?****Solutions –** Concerned event = 1(One head)

Total Event = 2**⇒ P(E) = 1/2**__ II. Two Coins or One Coin Twice:__When two coins are tossed together or one coin is tossed in twice then following outcomes can be obtained:

**Here, ‘H’ = Head; ‘T’ = Tail.**

**Question: What is the Probability of getting at most one head on tossing a coin?****Solutions –** At most one head means there can be 0 head or there can be 1 head.

So, Concerned Event = **3 {(H,T) (T,H) (T,T)}**

Total Events = 4**⇒ P(E) = 3/4**

**III. ****Three Coins or One Coin Thrice:**

When three coins are tossed together or one coin is tossed in thrice then following outcomes can be obtained:

Here, ‘H’ = Head; ‘T’ = Tail.**(HHH), (HHT), (HTH), (THH), (HTT), (THT), (TTH), (TTT)**

Here, (HTH) shows that coin 1 has a head, coin 2 has a Tail while coin 3 has a head.

In such cases, Total Events = 8

**Question:** **Three unbiased coins are tossed. What is the probability of getting at least 2 heads?**

Here, at least heads means there can be 2 heads and 3 heads.

So, Concerned Events = **4 {(HHH), (HHT), (HTH), (THH)}**

Total Events = 8**⇒ P(E) =4/8 =1/2**

**3. Cards:**

There are four kinds of symbol used in playing cards. The etymology for different symbols is as below:**i) Spade** ⇒ ♠ ⇒ Black in color (13 in number)**ii) Club** ⇒ ♣ ⇒ Black in color (13 in number)**iii) Heart** ⇒ ♥ ⇒ Red in color (13 in number)**iv) Diamonds** ⇒ ♦ ⇒ Red in color (13 in number)

**Each of these 4 variants have 13 numbers each as 1, 2, 3 …. 10 and, Jack, Queen, King and Ace.** There are**1)** 26 red cards and 26 black cards.**2)** 4 cards each of 1, 2, 3 …. 10 and, Jack, Queen, King and Ace.**3)** 13 cards each of Spade, Heart, Club and Diamond.**So, in total there are 13 × 4 = 52 cards.**

**Types of questions asked:**

** 1. One card drawn:**In such types of question a card is drawn from the pack of cards.

**Here, the Total Events = 52**

**Question:** **What is the probability of getting a King of Spade or Queen of Heart in one draw?**

**Solutions –** Here, ‘King of Spade or Queen of Heart’ means that either the card can be the Spade King or Heart Queen. Clearly, there is only one King of Space and only one Queen of Heart.

So, Concerned Event = 2

Total Event = 52**⇒ P(E) = 2/52 = 1/26**

__2. More than One Card drawn__:

In such questions when more than One card is drawn we use the concept of Combination formula. For example the question below:

**Question – Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart is ____ ?**

Here, the ‘one spade’ card has to be drawn from 13 spade cards, **so its event = ^{13}C_{1}**

And, ‘one heart’ card has to be drawn from 13 heart cards,

**so its event =**

^{13}C_{1}**So, Concerned Event =**

^{13}C_{1}×^{13}C_{1}Total Events (as two cards are to be drawn from 52) =

^{52}C

_{2}

**So, P(E) = (**

^{13}C_{1}×^{13}C_{1})÷^{52}C_{2 }= (13 x 13) x 2 / 52 x 51 = 13 / 102**Question – ** **Two cards are drawn together from a pack of 52 cards. The probability that either both are red or both are Kings ____ ?**

Here, the ‘both red’ cards have to be drawn from 26 red cards, so its event =** ^{26}C_{2} **

Or, ‘both king’ cards have to be drawn from 4 King cards, so its event =

^{4}C

_{2}But there are

**And, they will be taken out of two red king cards only so they’ll be deducted from concerned events.**

__Two Red Kings which are common in both Red cards & King cards so they have been double counted__.**So, Concerned Event =**

^{26}C_{2}+^{4}C_{2 }–^{2}C_{2}Total Events (as two cards are to be drawn from 52) =

^{52}C

_{2}

**So, P(E) = (**

^{26}C_{2}+^{4}C_{2 }–^{2}C_{2})÷^{52}C_{2 }= [ (26 x 25) + (4 x 3) – 1 ] / 52 x 51 = 330 / 1326 = 55 / 221**4. Balls**:

In such questions, a bag contains certain balls and some ball(s) is(are) drawn.

__I. One ball drawn:__

**Question –** **In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked at random. What is the probability that it is neither red nor green?**

**Solutions –** Total Events = 8 (red) + 7 (blue) + 6 (green) = 21

Since, the selected ball has to be neither red nor green then it’d be Blue and blue balls are 7.

So, Concerned Event = 7 **⇒ P(E) = 7/21 = 1/3 **

**II. More than One ball drawn without replacement:**

**Question –** **A box contains 10 black or 10 white balls. The probability of drawing two balls of same colors?**

**Solution –** Total Events **(as 2 balls are drawn) = ^{20}C_{2}**

The balls drawn can either be

**both black color or both white color and for ‘OR’ event we add up the two events.**

**So, Concerned Event =**

^{10}C_{2}(if black) +^{10}C_{2}(if white)⇒ P(E) = (

^{10}C

_{2}+

^{10}C

_{2})÷

^{20}C

_{2 }= 10 x 9 + 10 x 9 / 20 x 19 = 9 / 19

**Question –** **A box contains 10 black and 10 white balls. The probability of drawing two balls of same colors?**

**Solution –** Total Events **(as 2 balls are drawn) = ^{20}C_{2}**

The balls drawn can be

**both black color or both white color. for ‘And’ event we multiply the two events.**

**First case – If both ball is white** –

**So, Concerned Event = ^{10}C0 (if black) x ^{10}C_{2} (if white)**

⇒ P(E) = (^{10}C0 x ^{10}C_{2})÷ ^{20}C_{2 }= 1 x 10 x 9 / 20 x 19 = 9 / 38

**Second case –** **If both ball is black **–

So, Concerned Event** = ^{10}C2 (if black) x ^{10}C0 (if white)**

**⇒ P(E) = ( ^{10}C2 x ^{10}C0)÷ ^{20}C_{2 }= 10 x 9 x 1 / 20 x 19 = 9 / 38**

__ ____Other__** Miscellaneous questions:**

**Qs 1. Four persons are chosen at random from a group of 3 men, 2 women and 4 children. What is the probability of exactly two of them being children?**

**Solution –** Total People = 3 + 2 + 4 = 9

Since, 4 people are chosen from 9 so Total Event = ^{9}C_{4}

And, since 2 of the chosen people have to be children so these 2 person have to be from 4 children so Concerned Event for children= ^{4}C_{2}And, other 2 people will be from Men & Women (3+2 = 5), their Concerned event = ^{5}C_{2}⇒ Concerned Event (for all 4 people) = ^{4}C_{2 }×^{5}C_{2}**⇒ P(E) = ( ^{4}C_{2 }×^{5}C_{2})_{ ÷ }^{9}C_{4 = (4 x 3 x 5 x 4) x (2 x3) / 9 x 8 x 7 x 6 = 10 / 21 }**

**Qs. 2. A and B give exam. Chance of husband’s selection is 1/7 and of wife’s selection is 1/5. Find probability of only one of them is selected.**

**Solution –** P( husband selecting) = 1/7

P (husband not selecting) = 1 – 1/7 = 6/7P( wife selecting) = 1/5

P (wife not selecting) = 1 – 1/5 = 4/5

Probability of only one of them is selected = P( husband selecting)× P (wife not selecting) + P( wife selecting)× P (husband not selecting) = (1/7 x 4/5) + (1/5 x 6/7) = 10/35 = 2/7.

**Thanks!**

**Tanushree Sharma**