## Probability questions for bank exams

**Probability questions shortcuts for bank exams:**In most of the

**bank PO exams**like

**IBPS PO,SBI PO,IBPS RRB PO and RBI officer**, under quantitative section 2 to 4 questions are asked from

**probability**section. Generally candidates take this section lightly and tend to loose 2,3, or 4 valuable marks.If you are ready to spend few hours in preparing probability section during your entire exam preparation period , you can come out of the exam hall with a smile.Here we are sharing important formulas and facts,

**easy way to solve probability questions for bank PO and Clerical examinations.**In clerical examinations only simple questions from probability sections are asked,while questions asked in bank PO examination demand some preparation.So,here we are sharing

**How to solve probability questions faster for bank examination**.We have included all

**important formulas of probability.**At the end you can also find some

**solved practice questions on probability**,which will be useful in your bank PO exam preparation

**Prerequisites For probability section preparation**

-Knowledge of basics of set theory

-permutation and combination

Read Shortcuts for permutation and combination for bank exams.

__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.

__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 possible 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.

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

- In tossing a coin
**S={H,T}** - If two coins are tossed
**S={HH,HT,TH,TT}** - In rolling a dice
**S={1,2,3,4,5,6}**

**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**

**Results on probability**

- P(S)=1
- 0≤P(E)≤1
- For any events A and B P(A∪B)=P(A)+P(B)-P(A∩B) ∪=Union,∩=Intersection
- P(A)=1-P(not A)

__Probability Practice questions for bank PO and Clerk examination__

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

*Question1**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

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

*Question2**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

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

*Question3**Solution*:n(S)=Number of ways two balls can be drawn from 10 balls(6 white +4 blacks)=

^{10}C

_{2}=(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=

^{6}C

_{2}+

^{4}C

_{2}=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)=

^{52}C

_{2}=(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 A∩B

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=

^{26}C

_{2}=325

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

^{4}C

_{2}=6

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

^{2}C

_{2}=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 '∪'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∩B)= P(A)+P(B)- P(A∪B)=1/1326

**: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?**

*Question5**Solution:*

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 %.

**: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?**

*Question6**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)=

^{2}C

_{2}=1

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

^{3}C

_{2}=3

Number of ways of selecting 1 red ball and 1 green ball n(E3)=

^{2}C

_{1}*

^{3}C

_{1}=2*3=6

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

^{7}C

_{2}=21

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

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

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