9th Class Mathematics Probability

Probability

Category : 9th Class

 

PROBABILITY

 

Learning Objectives

 

  • Probability
  • Playing Cards

 

Probability

 

A mathematically measure of uncertainty is known as probability. If there are 'a' elementary events associated with a random experiment and 'b' of them are favourable to event 'E':

 

  • Then the probability of occurrence of event E is denoted by P (E).

\[\therefore \,\,\,\,\,\,P(E)=\frac{b}{a}\] \[\Rightarrow \,\,0\le P\left( E \right)\le 1\]

  • The probability of non-occurrence of event E denoted by P(E) and is defined \[as\,\,\frac{a-b}{a}\].

\[\therefore \,\,\,\,\,\,\,P\,(\overline{E})\,=\frac{a-b}{a}=1-\frac{b}{a}=1-P(E)\]

  • \[\Rightarrow \,P\left( E \right)+P\left( \overline{E} \right)=1\]

 

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 outcome cannot be predicted is called a random experiment.

 

Example: Rolling an unbiased dice has all six outcomes (1, 2, 3, 4, 5, 6) known but exact outcome can be predicted.

Outcome: The result of a random experiment is called an outcome,

Sample Space: The set of all possible outcomes of a random experiment is known as sample space.

 

Example: The sample space in throwing of a dice is the set (1, 2, 3, 4, 5, 6).

Trial: The performance of a random experiment is called a trial.

 

Example: The tossing of a coin is called trial.

 

Event

 

An event is a set of experimental outcomes, or in other words it is a subset of sample space.

 

Example: On tossing of a dice, let A denotes the event of even number appears on top A: {2, 4, 6}.

Mutually Exclusive Events: Two or more events are said to be mutually exclusive if the occurrence of any one excludes the happening of other in the same experiment. E.g. On tossing of a coin is head occur, then it prevents happing of tail, in the same single experiment.

Exhaustive Events: All possible outcomes of an event are known as exhaustive events. Example:

In a throw of single dice the exhaustive events are six {1, 2, 3, 4, 5, 6}.

Equally Likely Event: Two or more events are said to be equally likely if the chances of their happening are equal.

 

Example: On throwing an unbiased coin, probability of getting Head and Tail are equal.

 

Playing Cards

 

  • Total number of card are 52.
  • There are 13 cards of each suit named Diamond, Hearts, Clubs and Spades.
  • Out of which Hearts and diamonds are red cards.
  • Spades and Clubs are black cards.

There are four face cards each in number four Ace, King, Queen and Jack.

 

Black Suit (26)

Red Suit (26)

Spade (13) & Club (13)

Diamond (13) & Heart (13)

 

  • Each Spade, Club, Diamond, Heart has 9 digit cards 2, 3, 4, 5, 6, 7, 8, 9 and 10.
  • There are 4 Honour cards each Spade, Club, Diamond, Heart contains 4 numbers of Honours cards Ace, King, Queen and Jack.

 

Commonly Asked Questions

 

  • In a through of a coin find the probability of getting a tail.

(a) \[\frac{1}{2}\]           (b) \[\frac{3}{2}\]

(c) \[\frac{1}{3}\]           (d) \[\frac{1}{4}\]

(e) None of these

 

Answer: (a)

Explanation: In this case sample space, S = {H, T}, Event E = {T}

\[\therefore \,\,\,P(E)\,=\frac{n(E)}{n(S)}=\frac{1}{2}\]

 

  • An unbiased die is tossed. Find the probability of getting a multiple of 2.

(a) \[\frac{1}{4}\]           (b) \[\frac{3}{2}\]

(c) \[\frac{1}{3}\]           (d) \[\frac{1}{2}\]

 

Answer (d)

Explanation: S = {1, 2, 3, 4, 5, 6}, Event E = {2, 4, 6} multiple of 2

\[\therefore \,\,\,P(E)\,=\frac{n(E)}{n(S)}=\frac{3}{6}=\frac{1}{2}\]

 

  • An unbiased die is tossed. Find the probability of getting a number less than or equal to 4.

(a) \[\frac{1}{3}\]           (b) \[\frac{2}{5}\]

(c) \[\frac{2}{3}\]          (d) \[\frac{1}{6}\]

(e) None of these

 

Answer: (c)

Explanation: Here Sample space S = {1, 2, 3, 4, 5, 6}, Event E = {1, 2, 3, 4} number- less than or equal to 4.

\[\therefore \,\,\,P(E)\,=\frac{n(E)}{n(S)}=\frac{4}{6}=\frac{2}{3}\]

 

  • What is the chance that a leap year selected randomly will have 53 Sundays?

(a)  \[\frac{2}{5}\]          (b) \[\frac{2}{7}\]

(c) \[\frac{1}{7}\]                       (d) \[\frac{2}{4}\]

(e) None of these

 

Answer: (b)

Explanation: A leap year has 366 days, out of which there are 52 weeks and 2 more days.

2 more days can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday),

(Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)

= n(S) = 7

So, (Sunday, Monday) and (Saturday, Sunday) = n (E) = 2, therefore chances that a leap year selected randomly will have 53 Sundays:

                        \[\therefore \,\,\,P(E)\,=\frac{n(E)}{n(S)}=\frac{2}{7}\]

  • What is the chance that a normal year selected randomly will have 53 Sundays?

(a) \[\frac{1}{7}\]           (b) \[\frac{2}{7}\]

(c) \[\frac{2}{8}\]           (d) \[\frac{2}{6}\]

(e) None of these

 

Answer: (a)

Explanation: A normal year has 365 days, out of which there are 52 weeks and 1 more day

So, extra day can be Sunday, Monday, Tuesday, Wednesday, Thursday Friday, Saturday

So, n(S) = 7, n(E) = 1

            \[\therefore \,\,\,P(E)\,=\frac{n(E)}{n(S)}=\frac{1}{7}\]

 

Other Topics

Notes - Probability


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