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Introduction to Probability - Calculate theoretical probability using sample space and event definitions

Grade 9CBSE

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

🔑Concepts

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An Experiment is an operation which can produce some well-defined outcomes. A Random Experiment is an experiment in which all possible outcomes are known and the exact outcome cannot be predicted in advance.

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The Sample Space (SS) is the set of all possible outcomes of a random experiment. For example, if a coin is tossed, S={H,T}S = \{H, T\}.

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An Event (EE) is a subset of the sample space. It represents a specific collection of outcomes we are interested in.

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The Theoretical Probability of an event EE, written as P(E)P(E), is calculated under the assumption that all outcomes of the experiment are equally likely.

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The probability of any event EE always lies between 00 and 11, inclusive: 0≤P(E)≤10 \le P(E) \le 1.

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An Impossible Event has a probability of 00 (e.g., getting a 77 on a standard six-sided die).

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A Sure Event (or Certain Event) has a probability of 11 (e.g., getting a number less than 77 on a standard die).

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For any event EE, the event 'not EE' is called its Complementary Event, denoted by Eˉ\bar{E}. The sum of their probabilities is always 11.

📐Formulae

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

P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)}

P(E)+P(Eˉ)=1P(E) + P(\bar{E}) = 1

P(Eˉ)=1−P(E)P(\bar{E}) = 1 - P(E)

💡Examples

Problem 1:

A fair die is thrown once. What is the probability of getting a prime number?

Solution:

The sample space is S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}, so n(S)=6n(S) = 6. The prime numbers in this set are 2,3,2, 3, and 55. Let EE be the event of getting a prime number. Then E={2,3,5}E = \{2, 3, 5\} and n(E)=3n(E) = 3. Using the formula: P(E)=n(E)n(S)=36=12P(E) = \frac{n(E)}{n(S)} = \frac{3}{6} = \frac{1}{2}

Explanation:

First, identify all possible outcomes (1 to 6). Then, identify the outcomes that satisfy the condition (prime numbers: 2, 3, 5). Divide the number of favorable outcomes by the total outcomes.

Problem 2:

A bag contains 55 red, 88 white, and 44 green marbles. One marble is taken out of the bag at random. What is the probability that the marble taken out is not red?

Solution:

Total number of marbles n(S)=5+8+4=17n(S) = 5 + 8 + 4 = 17. Let RR be the event that the marble is red. n(R)=5n(R) = 5. The probability of picking a red marble is: P(R)=517P(R) = \frac{5}{17} The event 'not red' is Rˉ\bar{R}. Using the complementary event formula: P(Rˉ)=1−P(R)=1−517=17−517=1217P(\bar{R}) = 1 - P(R) = 1 - \frac{5}{17} = \frac{17 - 5}{17} = \frac{12}{17}

Explanation:

To find the probability of an event 'not happening', you can either add up all other outcomes (white and green) or subtract the probability of the event happening from 1.

Problem 3:

Two coins are tossed simultaneously. Find the probability of getting at least one head.

Solution:

When two coins are tossed, the sample space is S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}, so n(S)=4n(S) = 4. Let EE be the event of getting 'at least one head'. The favorable outcomes are HH,HT,HH, HT, and THTH. So, n(E)=3n(E) = 3. P(E)=34P(E) = \frac{3}{4}

Explanation:

The term 'at least one' means one or more. In the case of two coins, this includes the outcomes with one head and the outcome with two heads.