# Probability - Class 10 : Notes

Notes for probability chapter of class 10 Mathematics. Dronstudy provides free comprehensive chapterwise class 10 Mathematics notes with proper images & diagram.

(1) In the experimental approach to probability, We find the probability of the occurrence of event by actually performing the experiment a number of times and adequate recording of the happening of event.

(2) In the theoretical approach to probability, we try to predict what will happen without actually performing the experiment.

(3) An outcome of a random experiment is called an elementary event.
For Example: Consider the random experiment of tossing coin. The possible outcome of this experiment are head(H) and tail (T). if we define ${E_1}$ = getting head(H), ${E_2}$ = getting tail (T)

Then, ${E_1}$ and ${E_2}$ are elementary associated with the experiments of tossing of a coin.

(4) An event associated to a random experiment is a compound event if it is obtained by combining two or more elementary events associated to the random experiment.
For Example: In a single throw of a die, the event “getting an even number” is a compound event as it is obtained by combining three elementary events, namely 2, 4, 6.

(5) An event A associated to a random experiment is said to occur if any one of the elementary events associated to the event A is an outcome.
For Example: Consider the random experiment of throwing an unbiased die. Let A denote the event “getting an even number ”. Elementary events associated to this event are 2, 4, 6. Now, suppose that in a trail the outcome is 4, then we say that the event A has occurred. In another trail , let the outcome be 3, then we say that the event A has not occurred.

(6) An elementary event is said to be favourable to a compound event A, if it satisfies the definition of the compound event.
For Example: Consider the random experiment of two coins are tossed simultaneously and A is an event associated to it defined as “getting exactly one head”. We say that the event A occurs if we get either HT or TH as an outcome. So, there are two elementary events favourable to the event A.

(7) If there are n elementary events associated with a random experiment and m of them are favourable to an event A, then the probability of happening occurrence of event A is denoted by $P(A)$ and is defined as the ratio ${m \over n}$
i.e., $P(A) = {m \over n}$
For Example: Let A denote the event “getting an even number”

Clearly , event A occurs if we obtain any one of 2, 4, 6 as an outcome.
Favourable number of elementary events=3
$P(A) = {3 \over 6} = {1 \over 2}$

(8) For any event A associated to a random experiment, we have
(i) $0 \le P(A) \le 1$ (ii) $\overline {P(A)} = 1 - P(A)$
Proof of (i):
$P(A) = {m \over n}$

$0 \le m < n$
$0 \le {m \over n} \le 1$
$0 \le P(A) \le 1$

Proof of (ii):
If $P(A)=1$, then A is called a certain event and A is called an impossible event, if $P(A)=0$.

If m elementary events are favourable to an event A out of n elementary events, then the number of elementary events which ensure the non-occurrence of A. i.e. the occurrence of ${\bar A}$ is $n - m$
$P(\overline A ) = {{n - m} \over n}$
$P(\overline A ) = 1 - {m \over n}$
$P(\overline A ) = 1 - P(A)$

(9) The probability of a sure event is 1.
For Example: sun is rising from the east. this is a sure event. so probability of sure event is 1

(10) The probability of an impossible event is 0.
For Example: suppose the sun is rising from the west. this event is impossible event so probability of impossible event is always 0.

(11) The sum of the probabilities of all the outcomes (elementary events) of an experiment is 1.
For Example: Suppose in experiment of tossing coin 10 times 6 time head appear and 4 times tails appear.

So probability of getting head is $P(H) = {m \over n}$
where m is number of time head appear And n is number of time tossing coin.
So, $P(H) = {6 \over {10}} = 0.6$.
Now probability of getting tail is given by $P(T) = {m \over n}$
where m is number of time tail appear and n is number of time tossing coin.
So, $P(T) = {4\over {10}} = 0.4$.
So total probability of this experiment is given by $P(A) = P(H) + P(T)$. $P(A) = 0.6 + 0.4$ $P(A) = 1$.

Hence sum of probability of all outcomes of an experiment is always 1.

• Joyce mary.k

No

• Nice notes for board exam

• shubham

nice

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