[Maths Class Notes] on CBSE Class 10 Maths Chapter 15 – Probability Formula Pdf for Exam

Probability Class 10 Terminologies

There are some terminologies that are associated with probability formulas. Class 10 probability formulas contain major terminologies like:

• Experiment: It is any situation or a phenomenon like tossing a coin, playing cards, rolling dice, etc.

• Outcome: The result of an event after performing an experiment like the number appearing on dice after rolling, the side of the coin after flipping, and a card drawn out from a pack of well-shuffled cards, etc.

• Event: The combination of all possible outcomes of an experiment like getting king in a pack of well-shuffled cards, getting head or tail on a tossed coin, getting even or odd numbers on dice, etc.

• Sample Space: The set of all possible results or outcomes is sample space, Like getting Head or tail while flipping a coin.

In Probability Class 10 formulas main points are:

Probability = (No. of  Favourable outcome divided by Total)/(No. of possible outcomes)

P = n(E)/n(S)

Where n(E) = Number of favourable outcomes.

 n(S) = Total number of outcomes.

Types of Probability

There are three types of probabilities:

• Theoretical Probability

• Experimental Probability

• Axiomatic Probability

Theoretical Probability

It is based on the possible chances of something happening. The theoretical probability is based on the reasoning behind the probability. Eg: if a coin is tossed once, the theoretical probability of getting a head or a tail will be ½.

Experimental Probability

Experimental probability is directly based on the basis of observations of any experiment. It can be calculated on the basis of the number of possible outcomes by the total number of trials. Eg: if a coin is tossed 8 times and the heads are recorded 6 times then, the experimental probability for heads is 6/8 or, 3/4.

Axiomatic Probability

In this type of probability, a set of rules or called axioms are set applied to all types. These axioms are known as Kolmogorov’s three axioms. With this approach of probability, the chances of occurrence or non-occurrence of the events can be quantified.

Questions on Probability Class 10 Formulas

Probability Class 10 Formulas consist of three types main types questions which are discussed below:

Tossing a Coin

When a coin is tossed once, there are only two possible outcomes Either head (H) or tails (T).

So the total number of possible outcomes is 2 (Either head or tails).

So, we can say that the probability of getting H is ½ and the probability of getting T is ½.

When two coins are tossed simultaneously, the sample space is given by S = {HH, HT, TH, TT} where, H is the appearance of the Head, and T is the appearance of the Tail on the coin.

So, the number of elements in the sample space = the total number of outcomes is given by the following: n (S) = 4.

When we have three coins tossed simultaneously the possibility of outcomes are : (HHH) or (HHT) or (HTH) or (THH) or (HTT) or (THT) or (TTH) or (TTT) respectively; where H is denoted for heads and T is denoted for the tails. So, the total number of outcomes is 23 = 8 and so on.

Throwing Dice

If a single die is thrown once, there will be only six possible outcomes: 1, 2, 3, 4, 5, 6.

And the probability of getting 1 comes out to be 1/6.

Similarly, the probability of getting 2, 3, 4, 5, 6 is 1/6.

Let’s understand it with an example:

Example: What is the probability of getting an even number on rolling dice once.

Solution: Possible outcomes when dice is thrown  = {1, 2, 3, 4, 5, 6}

Event (E) = {2, 4, 6}

So, sample space = 6 and events = 3

Putting this in the probability formula, we get:

P(E) = 3/6 = 1/2 = 0.5

This concludes that the chances of getting an even number while rolling dice are 0.5.

When two dice are rolled simultaneously or Probability of rolling one dice twice with the six-sided dots such as 1, 2, 3, 4, 5, and 6 dots in each die can be 62 = 36 because each die has 1 to 6 numbers on its faces. The possible outcomes are given in the below table.

(i) The outcomes like (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), and (6, 6) coming in pairs are called doublets.

(ii) The pair (1, 3) and (3, 1) will be termed as different outcomes.

In the same way, if we have 3 dice the possible outcomes will be 63 = 216.

Playing Cards

In a pack or deck of playing cards, there is a total of 52 cards, they all are divided into 4 suits of 13 cards each spade ♠, hearts ♥, diamonds ♦, clubs ♣. Cards of Spades and clubs are black in colour while Cards of hearts and diamonds are red in colour. So, we can say that there are 26 red cards and 26 black cards.

The card in each suit is subdivided as ace, king, queen, jack or knaves, 10, 9, 8, 7, 6, 5, 4, 3, and 2. King, Queen, and Jack (or Knaves) are the face cards. So, it can be concluded easily that there are 12 face cards in total in the deck of playing cards.

Example: From a pack of well-shuffled cards, a card is drawn at random. Find the probability of getting:

(i) ‘2’ of spades.

(ii) A jack.

(iii) A king of red colour.

(iv) A card of diamond.

(v) A king or a queen.

Solution: As we know that in a playing card there is a total of 52 cards and hence the possible outcome is also 52.

And we know that probability =  (Number of favorable outcomes)/(Total number of possible outcome)

So solving the question:

(i) ‘2’ of spades:

The number of favourable outcomes i.e as discussed above ‘2’ of spades is 1 out of a total of 52 cards.

So, the probability of getting ‘2’ of spade = 1/52.

(ii) A jack

The number of favourable outcomes i.e. ‘a jack’ is 4 out of 52 cards.

So, the probability of getting ‘a jack’ = 4/52 = 1/13.

(iii) A king of red colour.

The number of favourable outcomes i.e. ‘a king of red colour’ is 2 out of a total of 52 cards.

So, the probability of getting ‘a king of red color’ = 2/52 = 1/26.

(iv) A card of diamond.

The number of favourable outcomes i.e. ‘a card of diamond’ is 13 out of 52 cards as discussed above. So, the probability of getting a card of diamond’= 13/52 = ¼.

(v) A king or a queen.

The total number of the king is 4 out of 52 cards as discussed above.

And the total number of the queen is 4 out of 52 cards as discussed above.

A number of favourable outcomes i.e. ‘a king or a queen’ can be calculated as 4 + 4 = 8 out of 52 cards. So, the probability of getting a king or a queen’= 8/52 = 2/13

Bayes’ Theorem

It is defined as a way of finding out probability when we know certain other probabilities.

The formula of which is given below:

[P(A|B) = frac{P(A) P(B|A)}{P(B)}]

This tells us how often A happens given that B happens which can be written as P(A | B),

How often B happens given that A happens can be written as P(B | A)

And how likely A is on its own can be written as P(A).

And how likely B is on its own can written as P(B)

Example: Dangerous fires are very rare around 1% but the smoke is fairly common around 10% due to barbecues, and 90% of dangerous fires make a smoke.

Solution: We can calculate the probability of dangerous Fire when there is Smoke by Bayes theorem:

[P(Fire|Smoke) = frac{P(Fire) P(Smoke|Fire)}{P(Smoke)}]

[= frac{1% times 90%}{10%}]

= 9%

We have completed probability formulas for Class 10, and maximum questions can be covered by practising more questions from all these topics.

Conclusion

Here, we have discussed all formula of probability Class 10, Probability is the most scoring topic in Class 10 and questions are very basic and simple, Student should focus mostly on calculating the sample space, Eg: If one coin is tossed once the sample space is 2 If 2 coins are tossed simultaneously sample space is 4 and so on.  Similarly, If a dice is thrown once sample space is 6 and if it is thrown twice sample space is 36. Try to cover probability all formulas Class 10 to secure good marks in the board examination.

Always remember If the question is based on coins sample space is (2)1 = 2 (Single coin) (2)2 = 4 (When there are two coins). Similarly with coins (6)1 = 6 (When dice is thrown once) and (6)2 = 36 (When there are two dice).