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Probability - Probabilities of equally likely outcomes

Grade 11CBSE

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

🔑Concepts

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A sample space SS is a set of all possible outcomes of a random experiment. If every outcome in SS has the same chance of occurring, they are called equally likely outcomes.

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If n(S)n(S) is the total number of elementary outcomes in a finite sample space SS and n(E)n(E) is the number of outcomes favorable to an event EE, then the probability of event EE is defined as P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)}.

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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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For any event EE, the probability of the complement event E′E' (event 'not EE') is given by P(E′)=1−P(E)P(E') = 1 - P(E).

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An event EE is called a sure event if P(E)=1P(E) = 1, and an impossible event if P(E)=0P(E) = 0.

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For any two events AA and BB, the probability of the union is P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B). If AA and BB are mutually exclusive, P(A∩B)=0P(A \cap B) = 0.

📐Formulae

P(E)=Number of outcomes favorable to ETotal number of possible outcomes=n(E)n(S)P(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of possible outcomes}} = \frac{n(E)}{n(S)}

P(E′)=1−P(E)P(E') = 1 - P(E)

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

P(S)=1,P(ϕ)=0P(S) = 1, \quad P(\phi) = 0

💡Examples

Problem 1:

A bag contains 55 red balls and 77 black balls. If one ball is drawn at random, what is the probability that it is a red ball?

Solution:

P(Red)=512P(\text{Red}) = \frac{5}{12}

Explanation:

The total number of balls in the bag is n(S)=5+7=12n(S) = 5 + 7 = 12. The number of red balls is n(E)=5n(E) = 5. Since each ball is equally likely to be drawn, the probability is n(E)n(S)=512\frac{n(E)}{n(S)} = \frac{5}{12}.

Problem 2:

Two dice are thrown simultaneously. Find the probability of getting a sum of 88.

Solution:

P(Sum is 8)=536P(\text{Sum is } 8) = \frac{5}{36}

Explanation:

When two dice are thrown, the total number of outcomes is n(S)=6×6=36n(S) = 6 \times 6 = 36. The outcomes favorable to the sum being 88 are E={(2,6),(3,5),(4,4),(5,3),(6,2)}E = \{(2,6), (3,5), (4,4), (5,3), (6,2)\}. Thus, n(E)=5n(E) = 5. The probability is P(E)=n(E)n(S)=536P(E) = \frac{n(E)}{n(S)} = \frac{5}{36}.

Problem 3:

Calculate the total number of ways to choose 22 cards from a deck of 5252 if order does not matter.

Solution:

52×512×1=1326\begin{array}{r} 52 \times 51 \\ \hline 2 \times 1 \end{array} = 1326

Explanation:

This uses the combination formula nCr=n!r!(n−r)!^nC_r = \frac{n!}{r!(n-r)!}. For choosing 22 cards from 5252, we calculate 52C2=52×512×1=1326^{52}C_2 = \frac{52 \times 51}{2 \times 1} = 1326.