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# Probability - A Theoretical Approach

Course

#### description

• Classical Definition of Probability
• Type of Event - Impossible and Sure Or Certain
• assume that all the experiments have equally likely outcomes, impossible event, sure event or a certain event, complementary events,

#### notes

Probobility means chances or possibilities of happening something.

1) "P(E)"= "No. of fovourable outcomes"/ "Total No. of possible outcomes"

Where, P(E) is read as probability of an event, fovourable outcomes are the outcomes that we are asked find in the question, total no. of possible outcomes is the total possible result.

Example1 - "P(H)"= "Probability of getting head"= 1/2

"P(T)"= "Probability of getting tail"= 1/2

Example2 - "P(S)"= "Probability of getting 5 on a die"= 1/6

2) Impossible Event- Some events can never take place, they are known as impossible events. The probablity of impossible events is always 0.

Example1 - Probability of getting 8 on a die "P(E)"= 0/6= 0

Example2 - Probability of sun revolving around the earth P(E)= 0

3) Sure Event- This are events which will definetly happen. The probablity of sure events is always 1.

Example- Probability of getting less than 7 on a die

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

4) Complementary Events- The sum of probablities of events that will happen and the one that will not happen is known as complementary events. The sum of complemetary event is always 1. P(E)+ P(not E)=1.

P(not E)= 1- P(E)

Example- Probability of getting 3 on a die "P(E)"= 1/6

"P(E)"+ "P"("not E")= 1

1/6+ "P"("not E")= 1

"P"("not E")= 1- 1/6

"P"("not E")= 5/6

5) Note-

i) Probability for any event- 0 greater than equal to P(E) greater that equal to 1.

ii) Probability can neither be negative nor be greater than one.

iii) An event having only one outcome of the experiment is called an elementary event. The sum of the probabilities of all the elementary events of an experiment is 1.

iv) Sum of all the prbabilities of some events is one i.e

P(E_1)+ P(E_2)+ P(E_3)+ .......= 1

Example- P(E_1)= 1/3, P(E_2)= 1/4, P(E_3)= ?

Solution- P(E_1)+ P(E_2)+ P(E_3)= 1

1/3+ 1/4+ P(E_3)= 1

P(E_3)= 1- 7/12

P(E_3)= 5/12

6) Coins- In the case of coins we will toss upto three coins. Sample space means number of possible outcomes.

i) One coin- Sample space= Head(H), Tail(T). Total number of outcomes are 2.

ii) Two coins- Sample space= HH, HT, TH, HH. Total number of outcomes are 4.

iii) Three coins- Sample space= HHH, HHT, HTH, THH, TTT, TTH, THT, HTT. Total number of outcomes are 8.

7) Die- Here we will the sample space upto two dice.

i) One die- sample space= 1,2,3,4,5,6. Total number of outcomes are 6.

ii) Two dice- sample space= (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Total number of outcomes are 36

Doublet, means chances of getting same numbers on both the dice- sample space= (1,1) (2,2) (3,3) (4,4) (5,5) (6,6).

Total number of outcomes are 6.

8) Playing cards- The total numbers of cards in a pack are 52, and the number of suit i.e number of types are 4.

The 4 suits are spade, club, heart and diamond. There are only two colours in a pack of cards that are red and black. While spade and club are always black and heart and diamond are always red.

Out of total 52 cards 26 cards are red and other 26 are black. There are 13 cards of each suit. Eg. Spade will have cards like 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king), A (ace), every suit will all this cards.

Face cards or picture cards are the cards which have face on it i.e like Jack, Queen and King. There are in total 12 face cards. Because every suit will have one jack, one queen and one king, that's why there will be 12 face cards.

And if it is asked that in a pack of cards how many red face cards we have? Then the answer wil be 6. There are 6 red face cards and 6 black face cards.

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Probability part 2 (Classical Approach) [00:08:41]
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