Students should go through these JAC Class 9 Maths Notes Chapter 15 Probability will seemingly help to get a clear insight into all the important concepts.

### JAC Board Class 9 Maths Notes Chapter 15 Probability

**Probability:**

Theory of probability deals with measurement of uncertainty of the occurrence of some event or incident in terms of percentage or ratio.

→ Sample Space: Set of all possible outcomes.

→ Trial: Trial is an action which results in one of several outcomes.

→ An experiment: An experiment is any kind the of activity such as throwing a die, tossing a coin, drawing a card. The different possibilities which can occur during an experiment. e.g. on throwing a dice, 1 dot, 2 dots, 3 dots, 4 dots, 5 dots, 6 dots can occur.

→ An event: Getting a ‘six’ in a throw of dice, getting a head, in a toss of a coin.

→ A random experiment: is an experiement that can be repeated numerous times under the same conditions.

→ Equally likely outcomes: The outcomes of a sample space are called equally likely if all of them have same chance of occurring.

→ Probability of an event A: Written as P(A) in a random experiment and is defined as –

P(A) = \(\frac{\text { Number of outcomes in favour of A }}{\text { Total number of possible outcomes }}\)

**Important Properties:**

(i) 0 ≤ P(A) ≤ 1

(ii) P (not happening of A) + P (happening of A) = 1

or, P(\(\bar{A}\)) + P(A) = 1

∴ P(\(\bar{A}\)) = 1 – P(A)

Probability of the happening of A = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}}\)

Probability of not happening of A (failing of A) = \(\frac{\mathrm{n}}{\mathrm{m}+\mathrm{n}}\)

where event A can happen in m ways and fail in n ways. All these ways being equally likely to occur.

**Problems of Die:**

A die is thrown once. The probability of

→ Getting an even number in the throwing of a die: the total number of outcomes is 6. Let A be the event of getting an even number then there are three even numbers 2, 4, 6.

∴ Number of favourable outcomes = 3.

P(A) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)

= \(\frac{3}{6}=\frac{1}{2}\)

→ Getting an odd number: Total no. of outcomes = 6,

favourable outcomes = 3 i.e. {1, 3, 5}

∴ P(A) = \(\frac{3}{6}=\frac{1}{2}\)

→ Getting a natural number P(A) = \(\frac{6}{6}\) = 1

→ Getting a number which is multiple of 2 and 3 = \(\frac{1}{6}\)

→ Getting a number ≥3 i.e. {3, 4, 5, 6},

P(A) = \(\frac{4}{6}\) = \(\frac{2}{3}\)

→ Getting a number 5 or 6, P(A) = \(\frac{2}{6}\) = \(\frac{1}{3}\)

→ Getting a number ≤5 i.e. {1, 2, 3, 4, 5},

P(A) = \(\frac{5}{6}\)

Problems Concerning Drawing a Card:

- A pack of 52 cards
- Face cards (King, Queen, Jack)