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Probability - Probability of event 'not A'

Grade 11CBSE

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

🔑Concepts

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Complementary Event: For any event AA associated with a random experiment, the event 'not AA' is called the complement of event AA. It is denoted by A′A', A‾\overline{A}, or AcA^c.

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The event A′A' consists of all outcomes in the sample space SS that are not in event AA. In set notation, A′=S−AA' = S - A.

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Mutual Exclusivity: Event AA and its complement A′A' are mutually exclusive and exhaustive events, meaning A∩A′=ϕA \cap A' = \phi and A∪A′=SA \cup A' = S.

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The sum of the probability of an event and the probability of its complement is always equal to 11.

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The probability P(A′)P(A') always lies in the range [0,1][0, 1].

📐Formulae

P(A′)=1−P(A)P(A') = 1 - P(A) Or equivalently: P(A)+P(A′)=1P(A) + P(A') = 1

P(not A)=n(S)−n(A)n(S)P(\text{not } A) = \frac{n(S) - n(A)}{n(S)}

P(S)=1,P(ϕ)=0P(S) = 1, P(\phi) = 0

💡Examples

Problem 1:

The probability that it will rain tomorrow is 0.070.07. What is the probability that it will not rain tomorrow?

Solution:

Let AA be the event that it rains tomorrow. We are given P(A)=0.07P(A) = 0.07. We need to find P(A′)P(A'). P(A′)=1−P(A)P(A') = 1 - P(A) P(A′)=1−0.07P(A') = 1 - 0.07 1.00−0.070.93\begin{array}{r} 1.00 \\ -0.07 \\ \hline 0.93 \end{array} Therefore, P(A′)=0.93P(A') = 0.93.

Explanation:

Using the complement rule, the probability of an event not occurring is 11 minus the probability of the event occurring.

Problem 2:

A card is drawn from a well-shuffled deck of 5252 cards. Find the probability that the card drawn is not an Ace.

Solution:

Total number of outcomes n(S)=52n(S) = 52. Let EE be the event of drawing an Ace. There are 44 Aces in a deck, so n(E)=4n(E) = 4. P(E)=n(E)n(S)=452=113P(E) = \frac{n(E)}{n(S)} = \frac{4}{52} = \frac{1}{13} The event 'not an Ace' is E′E'. P(E′)=1−P(E)P(E') = 1 - P(E) P(E′)=1−113P(E') = 1 - \frac{1}{13} P(E′)=13−113=1213P(E') = \frac{13 - 1}{13} = \frac{12}{13}

Explanation:

First, calculate the probability of the event occurring, then subtract it from 11 to find the probability of the complement.

Problem 3:

In a single throw of a die, what is the probability of not getting a multiple of 33?

Solution:

Sample space S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}, so n(S)=6n(S) = 6. Let AA be the event of getting a multiple of 33. A={3,6}A = \{3, 6\}, so n(A)=2n(A) = 2. P(A)=26=13P(A) = \frac{2}{6} = \frac{1}{3} Probability of not getting a multiple of 33 is P(A′)P(A'): P(A′)=1−P(A)P(A') = 1 - P(A) P(A′)=1−13=23P(A') = 1 - \frac{1}{3} = \frac{2}{3}

Explanation:

Identify the outcomes favoring the event, find its probability, and apply the formula for the complementary event.