Probability notes

Quick Study Notes on Probability - Part I

Hello Readers,

Since you all know that Mathematics is one of the most important part of competitive exams, here we posting Some Quick Study Notes on Probability. We hope that they will help you all in the upcoming competitive exams like SBI PO 2015, Insurance Exams and IBPS CWE 2015. Hope you all like the post..

Remember points:

(1) Any value of probability is not more than – 1

O <  P (E) < 1

(2) Formulae of probability

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

Where: 

P(E) = Probability of an events.

n(E) = Total No. of required events.

n(S) = Total No. of possible events.

Probability divided into 4 parts

• Coin based problems
• Dice based problems
• Balls based problems
• Card based problems

1. Coin based problems:

I. Single – Coin

1. When a single coin throw at random them find the probability –

• a getting head
• a getting tell

1. Single coin are two face

{H, T} – Two events.

P (E) = n(E) / n(S)
= 1/2

1. 1/2 (Short Trick)

A getting – tail

2. 1/2

II. Double Coin

(1) Two coin are throw at random, then find the probability.

(i) A getting – Head

(ii) getting at least one head

(iii) both are head are getting

Solution:

4 events are required

Sample space – {HT< TH, HH, TT}

2\4 = 1\2

2. At least one head:

Events taken – {HT, TH, HH}

Þ 3/4

3. Both are Head

{HH}

Þ 1/4

No. of events – Short trick = 2ⁿ

Where n – no. of coin

2. Dice based problems:

I. Single dice:

When single dice throw at random find the probability that a getting even no.

Sample space: {1, 2, 3, 4, 5, 6}

Events : {2, 4, 6}

3/6 = 1/2 Þ Short trick

2. Double Dice 

When two dice thrown are random then find the probability that the sum on the top face of both the dice will be more than 9

Following are cases:

(5, 5) (6, 4) (4, 6) (6, 5) (5, 6) (6, 6)

So the total no. required events – 6

The total no. of possible events – 36

Hence probability – 6/36 = 1/6 

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