PROBLEMS ON PROBABILITY:PART 2-HOW TO SERIES
Hi Bankersdaily Aspirants,
Learning is a step by step process and you should gradually progress in that to ace the process. As you all know we have started a new How to Series to help with your preparations for various exams like IBPS PO , IBPS RRB Scale I officer , IBPS RRB Office Assistant , IBPS Clerk , IBPS SO , SBI PO , SBI Clerk , SBI SO and many other exams in this type.
Friends as I always say to ace in the exam we don’t want to get Cent Percentage, we just want to get the cut off mark in the exam.It is not that much difficult to get the cut off mark.There is certain topic in the Aptitude Section which we can solve in less than a minute & PROBABILITY is one among that.If you have practiced this topic then you can solve this with in a minute.We had already discussed some of the types of questions in Previous Article(Probability Part 1).Today we will discuss another set of question in this article.Before that have a glance of Part 1
BASED ON CARDS:
Friends in the last Article we had discussed problems on cards with the Keyword Both.In this Article we will discuss problems based on the Keyword
No, One from one type & Another from another type
13 Cards in Diamond,Heart,Spade & Club Contains:
Numbered Card:2,3,4,5,6,7,8,9,10(Totally 9 Numbered Cards)
Face Card:There are 4 Face Card they are (ACE,KING QUEEN & JACK)
9+4=13 CARDS in each Suit.
Now let us look into the Problem.
1)When 1 card is drawn at Random,What is the Probability that there will be NO King Card?
2)When three cards are drawn,What is the Probability that one card is Ace,one Card is King & one card is Jack?
Friends shall we solve together,
3 Cards are drawn,Total Possibility=52C3=52*51*50/1*2*3
Now,let us find the Possible Number of ways of getting 1 Ace,1 King & 1 Jack
There are 4 Ace in 52 Cards,
1 King in 52 Cards,
1 Jack in 52 Cards,
Now,we have to choose 1 card from each face.
4C1 * 4C1 * 4C1=4*4*4
Probability of getting 3 Cards=16/5525
BASED ON BALLS:
In this type,Questions will be asked in such a way that certain balls are drawn in different colours and after that there will be two conditions after this
1)They were Replaced.
2)They were not Replaced.
1)THEY WERE REPLACED:
A bag contains 8 Red & 10 Black Balls.If two draws of three balls are drawn randomly & the balls are replaced after the first draw then what is the probability that three balls are Red Balls at the first draw and Black Balls at the Second Draw?
Note:If “AND” is represented in the question then we have to “Multiply” the term.
If “OR” is represented in the question then we have to “Add” the term.
A Bag Contains 8 Red Balls & 10 Black Balls.Four Balls are Drawn one by one & they are not replaced.Find the Probability if the balls drawn are of Alternate of different Colours?(That is,if 1st ball is Red 2nd would be Black …,Vice Versa)
Friends I think this Article would help you to get an Idea About Probability.
If you have any Query Don’t Hesitate to Post your Query in the Comment Box we will Respond you as soon as possible