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Probability Formulae Simplified Revision Notes

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3.2.4 Probability Formulae

Probability is a fundamental concept in statistics, and understanding the key formulae is essential for solving problems related to random events. Here's a summary of the most important probability formulae, along with explanations and examples.

Basic Probability Formula

The probability of an event AA is given by:

P(A)=Number of favourable outcomesTotal number of possible outcomesP(A) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}
infoNote

Example: The probability of rolling a 44 on a 66-sided die:

P(Rolling a 4)=16P(\text{Rolling a 4}) = \frac{1}{6}

Complementary Probability

The probability of the complement of an event AA (i.e., the event not happening) is:

P(A)=1P(A)P(A') = 1 - P(A)
infoNote

Example: If the probability of rain tomorrow is 0.30.3, the probability that it won't rain is:

P(No Rain)=10.3=0.7P(\text{No Rain}) = 1 - 0.3 = 0.7

Addition Rule (Union of Two Events)

For any two events A A and BB , the probability that A A or BB (or both) occur is:

P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B)
infoNote

Example: If P(A)=:highlight[0.5]P(A) = :highlight[0.5] , P(B)=:highlight[0.4]P(B) = :highlight[0.4] , and P(AB)=:highlight[0.2]P(A \cap B) = :highlight[0.2] , then:

P(AB)=0.5+0.40.2=0.7P(A \cup B) = 0.5 + 0.4 - 0.2 = 0.7

Multiplication Rule (Intersection of Two Events)

For independent events AA and BB , the probability that both AA and B B occur is:

P(AB)=P(A)×P(B)P(A \cap B) = P(A) \times P(B)
infoNote

Example: If the probability of flipping a coin and getting heads is 12\frac{1}{2} , and the probability of rolling a 44 on a die is 16\frac{1}{6} , then the probability of getting heads and rolling a 44 is:

P(Heads and 4)=12×16=112P(\text{Heads and 4}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}

For dependent events AA and BB , the probability that both AA and BB occur is:

P(AB)=P(A)×P(BA)P(A \cap B) = P(A) \times P(B \mid A)

Where P(BA)P(B \mid A) is the conditional probability of B B given that AA has occurred.

infoNote

Example: In a deck of 5252 cards, the probability of drawing an Ace and then a King without replacement:

P(Ace and King)=P(Ace)×P(KingAce)P(\text{Ace and King}) = P(\text{Ace}) \times P(\text{King} \mid \text{Ace}) =452×451=162652= \frac{4}{52} \times \frac{4}{51} = \frac{16}{2652} =4663= \frac{4}{663}

Conditional Probability

The probability of event AA occurring given that event BB has already occurred is:

P(AB)=P(AB)P(B)P(A \mid B) = \frac{P(A \cap B)}{P(B)}
infoNote

Example: If there are 1010 red and 2020 blue marbles in a bag, and one red marble is drawn, the probability that a second marble drawn is red:

P(Red on first draw)=1030=13P(\text{Red on first draw}) = \frac{10}{30} = \frac{1}{3} P(Red on second drawRed on first draw)=929P(\text{Red on second draw} \mid \text{Red on first draw}) = \frac{9}{29}

Total Probability

If events B1,B2,,BnB_1, B_2, \ldots, B_n are mutually exclusive and exhaustive and AA is an event that can occur if any one of B1,B2,,BnB_1, B_2, \ldots, B_n occurs, then:

P(A)=P(AB1)+P(AB2)++P(ABn)P(A) = P(A \cap B_1) + P(A \cap B_2) + \ldots + P(A \cap B_n)

Or, using the conditional probability:

P(A)=P(B1)×P(AB1)+P(B2)×P(AB2)++P(Bn)×P(ABn)P(A) = P(B_1) \times P(A \mid B_1) + P(B_2) \times P(A \mid B_2) + \ldots + P(B_n) \times P(A \mid B_n)
infoNote

Example: If P(B1)=:highlight[0.4]P(B_1) = :highlight[0.4] , P(B2)=:highlight[0.6]P(B_2) = :highlight[0.6] , P(AB1)=:highlight[0.3]P(A \mid B_1) = :highlight[0.3] , and P(AB2)=:highlight[0.5]P(A \mid B_2) = :highlight[0.5] , then:

P(A)=0.4×0.3+0.6×0.5=0.12+0.3=0.42P(A) = 0.4 \times 0.3 + 0.6 \times 0.5 = 0.12 + 0.3 = 0.42

Bayes' Theorem

Bayes' Theorem allows the calculation of the probability of an event based on prior knowledge of conditions that might be related to the event. It is given by:

P(BiA)=P(Bi)×P(ABi)P(A)P(B_i \mid A) = \frac{P(B_i) \times P(A \mid B_i)}{P(A)}

Where BiB_i is one of the mutually exclusive and exhaustive events, and AA is the event for which we want to reverse the conditional probability.

infoNote

Example: If a medical test is 9595% accurate and 11% of the population has the disease, Bayes' Theorem can be used to calculate the probability that a person who tested positive actually has the disease.

Probability of "At Least One"

The probability that at least one event occurs is given by:

P(At least one)=1P(None)P(\text{At least one}) = 1 - P(\text{None})
infoNote

Example: The probability of rolling at least one 6 in two rolls of a die:

P(At least one 6)=1P(No 6s)P(\text{At least one 6}) = 1 - P(\text{No 6s}) =1(56×56)= 1 - \left(\frac{5}{6} \times \frac{5}{6}\right)=12536=1136 = 1 - \frac{25}{36} = \frac{11}{36}

Summary

These probability formulae are essential tools for calculating the likelihood of various outcomes in different scenarios. Understanding when and how to apply each formula allows for accurate analysis and decision-making in both theoretical and practical situations. The examples provided illustrate how these formulae are used in real-world contexts, helping to clarify their application.


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