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Conditional probability is when an event occurs only if certain conditions are met. This concept can be used to determine the likelihood of real-life events occurring under specific conditions, such as in weather forecasting we can calculate the probability of rain. In this topic, students will learn how conditional probability can be applied easily.
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Conditional probability is one of the commonly applied concepts in probability theory. In conditional probability, an event (A) is likely to occur given that another event (B) is occurring. Conditional probability can be applied when a student decides not to go on an excursion if it rains (5% probability), making the probability of going on the excursion 95%. In this example, the conditional probability is 5%.
Conditional probability can be mathematically expressed as A given
B or as P(A|B).
P(A|B) = P(A∩ B)/ P(B)
Here:
P(A|B): The probability of occurrence of A is dependent on the likelihood of B occurring.
P(A∩ B): Probability of both events A and B occurring at the same time.
P (B): Probability of the occurrence of event B.
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The three types of probability are conditional, joint, and marginal. Here, we will discuss a few key points that differentiate them:
In conditional probability, the probability of an event relies on another event in the past.
Here, Event A is drawing a four, and Event B is drawing a red card
If the probability of a card being drawn is red, the probability of drawing a four is:
P(A|B) = P(A∩ B)/ P(B) = 1/26 ÷1/2 = 2/26 = 1/13
A joint probability is the probability of two or more events occurring simultaneously. We assume that Event A is drawing a four (four of hearts or four of diamonds) and Event B is drawing a red card. Here, we determine the probability of drawing a card that is both red and a four, meaning both conditions need to be satisfied. It can be mathematically expressed as:
P(A∩ B) = 2/25 = 1/26
The marginal probability is the probability of a single event occurring, independent of any other event.
P (A) or P(B) = ∑ P(A∩ B)
For example: 70% like reading (A)
50% like writing (B)
30% like both reading and writing (A ∩ B)
The marginal probability of liking writing is:
P(B) = P (A ∩ B) + P (B ∩ AC)
To find the conditional probability, we use the following steps:
1. Identify the events:
2. Calculate the joint probability P (A ∩ B)
3. Determine the probability of the stated condition P(B).
4. Now, we use the formula : P(A|B) = P (A ∩ B) P(B)
P(A|B) = P(A)
P(B|A) = P(B)
These events satisfy the multiplication rule: P(A∩B) = P(A) × P(B)
The events that do not occur at the same time are referred to as mutually exclusive. For mutually exclusive events, the conditional probability will always equal zero.
P(A|B) = 0
P(B|A) = 0
Conditional probability plays a vital role in the prediction of various real-life situations based on conditions. Whether predicting the weather or solving complex real-life problems, conditional probability has numerous uses. Let’s look at a few examples:
Conditional probability is a significant concept in probability theory. Students can often make mistakes when determining conditional probability. These errors can be overcome with a proper understanding of the concept. Here are a few common mistakes along with some tricks to help you master them easily:
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Aden has a 60% chance of qualifying for an exam if they prepare and a 30% chance if they don’t. If the probability of studying is 80%, what is the probability that Aden qualifies for the exam?
The probability of Aden qualifying for the exam is 54% or 0.54.
We have:
A = qualifying for the exam
B = Preparing for the exam
The given probabilities are:
Probability of qualifying for the exam if he prepares
P (A|B) = 0.6
Probability of qualifying even without preparation:
P (A|Bc) = 0.3
Probability of preparing for the exam:
P(B) = 0.8
Probability of not preparing for the exam:
P(Bc) = 1 – P (B) = 1 – 0.8 = 0.2
Here, we use the law of total probability:
P(A) = P (A|B) P(B) + P(A|Bc) P(Bc)
Now, substitute the values into the equation:
P(A) = (0.6 × 0.8) + (0.3 × 0.2)
= 0.48 + 0.06
= 0.54
Therefore, we get 0.54 as the probability of Aden qualifying for the exam, which is equal to 54%.
In a class of 80 pupils, 50 learn Mathematics, 30 learn English, and 15 learn both Mathematics and English. If a pupil is selected at random and is found to be learning English, what is the probability that they also learn Mathematics?
The probability of a pupil who learns English and math at the same time is 0.5 or 50%.
Here,
A = Pupil learning mathematics
B = Pupil learning English
Now, we determine P(A|B):
P(A|B) = P(A∩ B)/ P(B)
We will now change the given values into probabilities:
P(A∩ B) = 15/80
P(B) = 30/80
Using the formula: P(A|B) = P(A∩ B)/ P(B)
= (15/80)/ (30/80) = 15/30 = 0.5
Therefore, the probability of a pupil who learns English and math at the same time is 0.5 or 50%.
On the turf, 70 children play badminton, and 30 play both badminton and basketball. If a child plays badminton, what is the probability that they also play basketball?
The probability of a child playing both badminton and basketball is 42.86% or 0.43.
We use the formula:
P(A|B) = P(A∩ B)/ P(B)
Here, we have:
P(A∩ B) = 30 (children playing both badminton and basketball)
P(B) = 70 (children playing badminton)
Substituting the values:
P(A|B) = 30/70 = 0.4286
Therefore, the probability of a child playing both badminton and basketball is 42.86% or 0.43.
A company manufactures 2,000 items daily. 600 come from Machine A, and 1400 from Machine B. Machine A makes 60 defective items, while Machine B makes 20 defective items. If an item is defective, what is the chance it came from Machine A?
The probability of a defective item from machine A is 0.75 or 75%.
We use the formula:
P(A|D) = P(A∩ D)/ P(D)
P (A) = Probability of an item from Machine A = 600/2000
P(B) = Probability of an item from Machine B = 1400/2000 = 0.7
P(D∣A) = Probability of a defective item from Machine A = 60/600 = 0.1
P(D∣B) = Probability of a defective item from Machine B = 20/1400 = 0.0143
Here, we use the total probability formula:
P(D) = P(D|A) P(A) + P(D|B) P(B)
P(D) = (0.1 0.3) + (0.0134 0.7)
= 0.03 + 0.01
= 0.04
Here, we apply Bayes’ theorem:
P(A|D) = P (D|A) P(A)/ P(D)
= (0.1 0.3)/0.04 = 0.03 /0.04 = 0.75
Therefore, the probability of a defective item from machine A is 0.75 or 75%.
A city records that 60% of youngsters drive above the speed limit. Among them, 25% are caught violating traffic rules. If a youngster is found violating traffic rules, what is the probability that they were also speeding?
0.75 is the probability.
Use the formula:
P(A|B) = P(A ∩ B)/ P(B)
A: The youngster was speeding
B : The youngster was caught
P(A) = 0.60
P(B|A) = 0.25
P(B) = 0.20
P (A∩ B) = P(B|A) P(A) = 0.25 0.60 = 0.15
P(A|B) = P(A ∩ B)/ P(B)
= 0.15/0.20 = 0.75
Therefore, if a youngster violates the traffic rules, the probability that they were speeding is 75%.
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Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref
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