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Probability - Simple Problems on Calculating Probability

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Theoretical Probability: The probability of an event EE is defined as the ratio of the number of outcomes favorable to EE to the total number of equally likely outcomes in the sample space. Visually, imagine a circle representing all possible outcomes; the event EE is a shaded slice of that circle.

Sample Space and Outcomes: The sample space is the set of all possible results of a random experiment. For example, when tossing two coins, the visual representation is a tree diagram with four branches: HHHH, HTHT, THTH, and TTTT. Each unique result is an 'outcome'.

Equally Likely Outcomes: Outcomes are equally likely if each has the same chance of occurring. For instance, a fair six-sided die, visually represented as a cube with dots 11 to 66, has a 16\frac{1}{6} probability for each face.

Range of Probability: The probability of any event EE is a real number such that 0P(E)10 \le P(E) \le 1. On a horizontal number line (a visual probability scale), 00 represents an 'Impossible Event' (far left), 0.50.5 represents an 'Even Chance' (middle), and 11 represents a 'Sure Event' (far right).

Complementary Events: For any event EE, the event 'not EE' (denoted as Eˉ\bar{E}) is called its complement. Together, they cover the entire sample space. Visually, if EE is a circle inside a rectangular box (the sample space), then Eˉ\bar{E} is everything inside the box that is outside the circle.

Elementary Events: An event having only one outcome of the experiment is called an elementary event. The sum of the probabilities of all the elementary events of an experiment is exactly 11.

Playing Cards Composition: A standard deck contains 5252 cards divided into 44 suits: Spades and Clubs (Black), and Hearts and Diamonds (Red). Each suit has 1313 cards. Visually, there are 1212 'face cards' in a deck (Kings, Queens, and Jacks), which are often depicted with illustrations rather than numbers.

📐Formulae

P(E)=Number of outcomes favourable to ENumber of all possible outcomesP(E) = \frac{\text{Number of outcomes favourable to } E}{\text{Number of all possible outcomes}}

0P(E)10 \le P(E) \le 1

P(E)+P(Eˉ)=1 or P(not E)=1P(E)P(E) + P(\bar{E}) = 1 \text{ or } P(\text{not } E) = 1 - P(E)

P(Sure Event)=1P(\text{Sure Event}) = 1

P(Impossible Event)=0P(\text{Impossible Event}) = 0

💡Examples

Problem 1:

A bag contains 33 red balls and 55 black balls. If a ball is drawn at random from the bag, what is the probability that the ball drawn is (i) red? (ii) not red?

Solution:

Step 1: Find the total number of outcomes. Total balls in the bag = 3+5=83 + 5 = 8. So, n(S)=8n(S) = 8. Step 2: Find the probability of drawing a red ball. Number of red balls n(R)=3n(R) = 3. P(R)=n(R)n(S)=38P(R) = \frac{n(R)}{n(S)} = \frac{3}{8} Step 3: Find the probability of not drawing a red ball using the complement rule. P(not R)=1P(R)=138=838=58P(\text{not } R) = 1 - P(R) = 1 - \frac{3}{8} = \frac{8 - 3}{8} = \frac{5}{8}

Explanation:

The problem uses the basic probability formula for the first part and the concept of complementary events for the second part to simplify the calculation.

Problem 2:

A single fair die is thrown once. Find the probability of getting (i) a prime number and (ii) a number lying between 2 and 6.

Solution:

Step 1: Identify the sample space. For a die, S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}, so n(S)=6n(S) = 6. Step 2: (i) Identify prime numbers in the sample space. Prime numbers are 2,3,52, 3, 5. Number of favorable outcomes n(E1)=3n(E_1) = 3. P(Prime)=36=12P(\text{Prime}) = \frac{3}{6} = \frac{1}{2} Step 3: (ii) Identify numbers between 2 and 6. The numbers are 3,4,53, 4, 5. Number of favorable outcomes n(E2)=3n(E_2) = 3. P(2<x<6)=36=12P(2 < x < 6) = \frac{3}{6} = \frac{1}{2}

Explanation:

This example requires identifying subsets of the sample space that satisfy specific conditions (primality and range) before applying the probability formula.