# 11th Class Mental Ability Probability Notes - Probability

Notes - Probability

Category : 11th 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).

$\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,P(E)=\frac{b}{a}\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,0\le P(E)\le 1$

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

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

• $\Rightarrow \,\,\,\,\,\,\,\,\,\,\,P(E)+P(\overline{E})=1$

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 unbiassed 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) Spades (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.

1. 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

Ans.     (a)

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

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

1. 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) $\frac{1}{4}$

(e) None of these

Ans.     (d)

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

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

1. 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

Ans.     (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.

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

1. 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

Ans.     (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.

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

1. 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

Ans.     (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

$\therefore \,\,\,P(E)=\frac{n(E)}{n(S)}=\frac{1}{7}$.

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##### Notes - Probability

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