Probability for competitive examinations

Probability questions for various competitive examinations

In most of the competitive exams like bank,IBPS,UPSC,SSC,MAT,CAT etc.., under quantitative section few questions are asked from probability section. Here we are sharing important formulas and facts,easy way to solve probability questions for competitive  examinations. .We have tried to include all important formulas of probability. At the end you can also find some solved practice questions on probability .

Prerequisites For probability section preparation

-Knowledge of basics of set theory

-permutation and combination

Probability-Important  formulas and facts

Experiment:An operation which can produce some well-defined outcomes is called an experiment.

Random Experiment:An experiment in which all possible outcomes are known and exact output cannot be predicted in advance is called a random experiments.For example, rolling a dice,tossing a coin,drawing a card from a well shuffled pack of cards etc.

Sample space: When we perform an experiment, then the set of all possible outcomes is called Sample Space.denoted by ‘S’.

Event:Any subset of Sample Space is called an Event

Probability of occurrence of an Event:Let S be the sample space and E be the Event,then probability of occurrence of E denote by P(E)

P(E)=n(E)/n(S)=Number of favourable outcomes/Number of possible outcomes

or 

P(E)=n(E)/n(S)=No of  Trials in which E  has happened / Total no of Trials

Some random experiments and their outcomes

Tossing a coin

When we toss a coin ,either Head(H) or a Tail(T) appears. If two coins are tossed simultaneously then possible outcomes are HH,HT,TH and TT.As number of coins increases possible outcomes also increases.

Rolling a Dice

A dice is a solid cube,having 6 faces,marked 1,2,3,4,5 and 6.When we roll a dice possible outcomes are 1,2,3,4,5 and 6.

If we roll two dice simultaneously posiible out comes are combination of two number (1,1)(1,2)(1,3)……………….(6,6)

Card is drawn from a pack of cards

A pack of card has 52 cards.

It has 13 cards of each suit, namely Spades, Clubs, Hearts and Diamonds.

Cards of Spade and Club are black.

Cards of Heart and Diamond are red cards.

There are 4 honours of each suit.These are Aces,Kings,Queens and Jack.These are called face cards.

Elementary Event

An event having only one out come. Sum of probabilities of all the elementary  events of an experiment is 1.

Eg. Rolling a dice, the  outcome of getting 2 is an elementary event. Similarly outcome of getting 1,3,4,5 or 6is also elementary events.

P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=1/6 +1/6 +1/6 +1/6 +1/6 +1/6 = 6/6 =1

Complementary Events

For any event E P(E)+P(E')=1, E and E' are complementary events 

here E' is read as  'E not'

)

Example:Probability of getting an 'ace' from a pack of 52 cards

Let E be the event of getting an 'ace'

Since there are 4 aces in a pack P(E)=4/ 52

Probability of not getting  an ace P(E')= 48/52

Here E and E' are complementary events.

P(E)+P(E')=1

Example:Tossing a coin , P(T)=1/2 , P(H)=1/2  and P(T)+P(H)=1

Here H & T are complementary events.

Mutually Exclusive Events(Disjoint Events)

Events That do not occur at the same time.

ie A ⋂ B=∅

Example:Tossing a coin, Head and Tail do not simultaneously.

Example:Rolling a die, 1 and 2 do not occur simultaneously.

Mutually Exclusice and Exhaustive events :

Events E1,E2,E3......En are mutually exclusive if

      Ei ⋂ Ej =∅ , where i≠j

and Exhaustive if 

     E1∪E2∪E3..........∪En=S

Conditional Probability

Conditional probability is  the possibility of an event or outcome happening, based on the existence of a previous event or outcome.

The conditional probability of an event E given the occurrence of the event F is given by 

P(E|F)

P(E|F)=P(E ⋂ F) / P(F)

Example: Ten cards numbered 1 to 10 placed in a box and mixed up thoroughly and then one card is drawn randomly. If it is known that number on the card is more that 3 what is the probability that it is an even number?

Ans: Let A be the event 'the number on the card is even' and B be the event ' the number on the card is more than 3'

The sample space or all possible outcomes 

        S={1,2,3,4,5,6,7,8,9,10}

        A={2,4,6,8,10}

        B={4,5,6,7,8,9,10}

Conventional Method:

Here favourable out comes are {4,6,8,10}

Since it is given that card drawn is more than 3, new sample space or all possible outcomes =B={4,5,6,7,8,9,10}

Required probability P(A|B)=(No of favourable outcomes/ No of Total outcomes) = 4/7

Using Formula P(A|B)=P(A ⋂ B) / P(B)

A ⋂ B={4,6,8,10}

P(A ⋂ B)=4/10

P(B)=7/10

Now P(A|B)=P(A ⋂ B) / P(B) = (4/10)/ (7/10)=4/7

ˉ 

Results on probability

• P(S)=1 (Probability of sample space)
• Probability of a sure event is 1
• Probability of an impossible event is "0"
• 0≤P(E)≤1
• For any events A and B
•  
• P(A)=1-P(not A)

Probability Practice questions

Question1:Two unbiased coin are tossed .What is the probability of getting at most one head?

Solution: Sample space (All [possible outcomes])S=(HH,HT,TH,TT)

Event(required outcomes)=(TT,HT,TH)

P(E)=n(E)/n(S)=3/4

Question2:Two dice are thrown simultaneously ,what is the probability of getting a total of 7?

Solutions:n(S)=6*6=36

E={(1,6),(6,1),(5,2),(2,5),(4,3),(3,4)}

n(E)=6

P(E)=6/36=1/6

Question3:A bag contains 6 white and 4 black balls. Two balls are drawn at random. Find the probability that they are of same colour.

Solution:n(S)=Number of ways two balls can be drawn from 10 balls(6 white +4 blacks)= ¹⁰C₂=(10*9)/(2*1)=45

n(E)=Number of ways of drawing 2 balls from 6 white balls or 2 balls from 4 black balls=⁶C₂+⁴C₂=21

P(E)=21/45=7/15

Question4:Two cards are drawn at random from a pack of 52 cards.What is the probability that either both are black or both are queens?

Solution:Total number of possible ways selecting 2 card from 52 cards is n(S)=⁵²C₂=(52*51)/(2*1)=1326

Let A be event of getting both black card.

B be event of getting both queens.

Event of getting two black queens AB

There are 26 black cards in a pack and 4 queens. Two queens are black

n(A)=Number of ways of selecting  2 blacks from 26 cards=²⁶C₂=325

n(B)=Number of ways of selecting 2 queens from 4 cards =⁴C₂=6

n(A∩B)=Chances of getting 2 black queens=²C₂=1

P(A)=325/1326

P(B)=6/1326

P(A∩B)=1/1326

P(A ∪ B)=P(A)+P(B)-P(A∩B)

                =(325/1326)+(6/1326)-(1/1326) = 330/1326 =55/221

Tip:Use union  U for ‘or’, ‘either’,’neither’

Use intersection  ∩ for ‘and’.

If in the above question ,question if changed like ‘chances of getting black queens’, then solution would be as follows.(ie getting two card that are back and queens)

= P(A)+P(B)- P(A=1/1326

Question5:A speaks truth in 75% cases and B in 80%  of cases. In what percentage of cases they contradict each other in narrating the same incident?

Probability that A is telling the truth P(A)=75/100=3/4

Probability that B is telling the truth P(B)=80/100=4/5

Probability that A is lying P(A’)=1-75/100 =25/100=1/4

Probability that B is lying P(B’)=1-80/100 =20/100=1/5

Contradiction means either of

             A telling truth and B lying

                            Or

       B telling truth and A lying

P(Contradiction)= P(A)*P(B’) + P(A’)*P(B)=(3/4) * (1/5) + (1/4)*(4/5)=7/20

7/20 is equal to 35 %.

Question6:A bag contains 2 red,  3 green and 2 blue balls.Two balls are drawn at random.What is the probability that none of the drawn ball is blue?

Solution:Possible out comes(E) or Favourable outcomes are

E1.2 red ball             or

E2.2 green ball            or

E3.1 red and 1 green

Number of selecting 2 balls from two red balls n(E)=²C₂=1

Number of ways of selecting 2 balls from 3 green balls n(E2)= ³C₂=3

Number of ways of selecting 1 red ball and 1 green ball n(E3)= ²C₁*³C₁=2*3=6

Sample space S=Number of ways of selecting 2 balls from 7(2+3+2) balls n(S)=⁷C₂=21

n(E)=n(E1)+ n(E2)+ n(E3)=1+3+6=10

Required probability P(E)=n(E)/n(S)=10/21