SOUTHERN BICOL COLLEGESMABINI ST. MASBATE CITY

COMPILATION OF THE DIFFERENT

KINDS OF PROBABILITY

(PROBABILITY)JUDE JEMMEL A. VENDEROBSED-III (Major in Math)

Engr. MARIA ROMINA P. ANGUSTIA INSTRUCTOR

TABLE OF CONTENTS

*History of Probability 1 *Complementary Probability 2 *Joint Probability 3

-Mutually Exclusive Event -Non-Mutually Exclusive Event 4

*Conditional Probability 5 -Dependent Probability -Independent Probability 6

*Repeated Trial Probability 7

History of probability

Probability has a dual aspect: on the one hand the likelihood of hypotheses given the evidence for them, and on the other hand the behavior of stochastic processes such as the throwing of dice or coins. The study of the former is historically older in, for example, the law of evidence, while the mathematical treatment of dice began with the work of Cardano, Pascal and Fermat between the 16th and 17th century.Probability is distinguished from statistics. (See history of statistics). While statistics deals with data and inferences from it, (stochastic) probability deals with the stochastic (random) processes which lie behind data or outcomes.

COMPLEMENTARY PROBABILITY-the probability of a given event not occurring, or of a

different event occurring that can only occur if the first event does not.

Formula:

P(A) + P(not A)=1

P(A)=1-P(not A)

P(not A)=1-P(A)

Example:

We draw a card from a given deck of cards. Find the probability of not getting a KING or a ACE.

solution:

There are 4 king and 4 ace so in total we do not acquire any of these 8 cards to be drawn.

So, probability of getting a king or a ace , P( king or ace ) =8/52=2/13

Hence the probability of not getting a king or a ace is given by:

P(not getting a king or ace)=1-2/13

=13/13-2/13

=11/13

=0.8461 or 84.61%

JOINT PROBABILITY-is the likelihood of more than one event occurring at the same time.

1.MUTUALLY EXCLUSIVE EVENT-without common outcomes

Formula:

P(A or B)=P(A)+P(B)

Example: A Deck of CardsSolution:

In a Deck of 52 Cards:

the probability of a King is 1/13, so P(King)=1/13 the probability of an Ace is also 1/13, so P(Ace)=1/13

 

When we combine those two Events:

The probability of a card being a King and an Ace is 0 (Impossible) The probability of a card being a King or an Ace is (1/13) + (1/13)

= 2/13

Which is written like this:

P(King and Ace) = 0

P(King or Ace) = (1/13) + (1/13) = 2/13

=0.1538 or 15.38%

2.NON-MUTUALLY EXCLUSIVE EVENT-with common outcomes

Formula:

P(A or B)=P(A)+P(B)-P(A and B)

Example:

Find the probability that a card from a deck will be either an King or Heart?

Solution:

Probability P(A) is P(King)=4/52

P(B) is P(heart)=13/52

Only way in a single draw to be King and Heart is King of Heart ; which is only one,

So, probability P(A and B) is P(King and Heart)=1/52

Therefore, the probability of event A or B is;

P(A or B)=P(A)+P(B)-P(A and B)

=P(King)+P(Heart)-P(King and Heart)

=4/52+13/52-1/52

=16/52 or 4/13, 0.3077, 30.77%

CONDITIONAL PROBABILITY-it is the probability of an event (A), given than another

(B) has already occurred.

2 types of conditional probability1.DEPENDENT PROBABILITY

-two events are dependent if the outcome or occurrence of the first affects the outcome or occurrence of the

second so that the probability is changed.

Formula:

P(B/A)=P(A and B)/P(A) or P(A and B)=P(A)xP(B/A)

Example: A card is chosen at random from a standard deck of 52 playing card. Without replacing it a second card is chosen . What is the probability that the first card chosen is a queen and the second card chosen is a jack?

Solution:

P(queen on first pick)  =  4 52

P(jack on 2nd pick given queen on 1st pick)  =

  4 51

P(queen and jack)  =  4

 ·  4

 =    16

 =    4

52

51

2652

663

=0.006033 or 0.6033%

2.INDEPENDENT PROBABILITY-independent events are not affected by previous events.

Formula:

P(A and B)=P(A)x P(B)

Example: A jar contains 3 red, 5 green, 2 blue and 6 yellow marbles. A marble is chosen at random from the jar. After replacing it, a second marble is chosen. What is the probability of choosing a green and then a yellow marble?

Solution:P(green)  =

  5 16

P(yellow)  =  6 16

P(green and yellow) = P(green) · P(yellow)

 =  5

 ·  6

16 16

 =   30 256

 =  15 128

=0.117 or 11.7%

REPEATED TRIALS PROBABILITY-probability that an event will occur exactly “r” times out

of “n” trials.

Formula:

P=nCr ( p)r(q)n−r

Where: p=probability of success q = probability of failure n=number of trials r=successful outcomes

Example: Coin A Fair coin is tossed 5 times. What is the probability of p – 1 / 2 𝑠𝑢𝑐𝑐𝑒𝑠𝑠 q - 1 / 2 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 n- 5 tosses/trials r- 3 heads(outcome)

Solution:

1. Substitute values to the formula: P=nC r (p )r (q )n−r

P(3h) = 5𝐶3 (0.5)3 (0.5)5−3 P(3h) = 𝟓/𝟏𝟔 or 0.3125 or 31.25%