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Probability - Probability — A Theoretical Approach

Grade 10CBSE

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

🔑Concepts

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Theoretical probability, also known as classical probability, is based on the assumption that outcomes of an experiment are equally likely.

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An event EE is a collection of some outcomes of the experiment. The probability of an event is denoted by P(E)P(E).

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The sum of the probabilities of all the elementary events of an experiment is 11.

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For any event EE, the value of P(E)P(E) always lies between 00 and 11 inclusive, i.e., 0≤P(E)≤10 \le P(E) \le 1.

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An impossible event has a probability of 00. For example, getting a number 77 when rolling a standard six-sided die.

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A sure event (or certain event) has a probability of 11. For example, getting a number less than 77 when rolling a standard die.

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Complementary Events: For any event EE, the event 'not EE' (denoted by Eˉ\bar{E}) is called its complement. The relationship is given by P(E)+P(Eˉ)=1P(E) + P(\bar{E}) = 1.

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In a deck of 5252 playing cards, there are 44 suits: Spades, Hearts, Diamonds, and Clubs. Each suit has 1313 cards. Spades and Clubs are Black; Hearts and Diamonds are Red. Face cards include King, Queen, and Jack (total 1212 face cards).

📐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)+P(Eˉ)=1P(E) + P(\bar{E}) = 1

0≤P(E)≤10 \le P(E) \le 1

💡Examples

Problem 1:

A die is thrown once. What is the probability of getting (i) a prime number; (ii) a number lying between 22 and 66; (iii) an odd number?

Solution:

Total possible outcomes when a die is thrown are {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}. Total number of outcomes =6= 6. (i) Prime numbers are {2,3,5}\{2, 3, 5\}. Number of outcomes =3= 3. P(prime)=36=12P(\text{prime}) = \frac{3}{6} = \frac{1}{2}. (ii) Numbers between 22 and 66 are {3,4,5}\{3, 4, 5\}. Number of outcomes =3= 3. P(between 2 and 6)=36=12P(\text{between 2 and 6}) = \frac{3}{6} = \frac{1}{2}. (iii) Odd numbers are {1,3,5}\{1, 3, 5\}. Number of outcomes =3= 3. P(odd)=36=12P(\text{odd}) = \frac{3}{6} = \frac{1}{2}.

Explanation:

We identify the total outcomes first, then identify the subset of outcomes that satisfy the specific condition to find the numerator for the probability formula.

Problem 2:

One card is drawn from a well-shuffled deck of 5252 cards. Find the probability of getting: (i) a king of red colour (ii) a face card (iii) a red face card.

Solution:

Total number of cards =52= 52. (i) There are 22 kings of red colour (King of Hearts and King of Diamonds). P(Red King)=252=126P(\text{Red King}) = \frac{2}{52} = \frac{1}{26}. (ii) There are 33 face cards in each suit (J, Q, K). Total face cards =3×4=12= 3 \times 4 = 12. P(Face card)=1252=313P(\text{Face card}) = \frac{12}{52} = \frac{3}{13}. (iii) Total red face cards =3= 3 (Hearts) +3+ 3 (Diamonds) =6= 6. P(Red Face card)=652=326P(\text{Red Face card}) = \frac{6}{52} = \frac{3}{26}.

Explanation:

The problem requires knowledge of the composition of a standard deck of cards. We divide the specific count of the required cards by the total count (5252).

Problem 3:

If P(E)=0.05P(E) = 0.05, what is the probability of 'not EE'?

Solution:

We know that P(E)+P(Eˉ)=1P(E) + P(\bar{E}) = 1. Given P(E)=0.05P(E) = 0.05. P(Eˉ)=1−P(E)P(\bar{E}) = 1 - P(E) P(Eˉ)=1−0.05=0.95P(\bar{E}) = 1 - 0.05 = 0.95

Explanation:

This uses the concept of complementary events where the sum of the probability of an event happening and not happening is always 11.