J and K are independent events. P(JK) = 0.3. Find P(J).
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![### Understanding Independent Events in Probability
**Problem Statement:**
J and K are independent events. \( P(J|K) = 0.3 \). Find \( P(J) \).
**Solution:**
To solve this, we need to understand the basic concepts of independent events in probability.
**Key Concept:**
For two events J and K to be independent, the occurrence of one event does not affect the probability of the occurrence of the other event.
Mathematically, two events J and K are independent if and only if:
\[ P(J \cap K) = P(J) \cdot P(K) \]
**Given:**
\[ P(J|K) = 0.3 \]
By the definition of conditional probability:
\[ P(J|K) = \frac{P(J \cap K)}{P(K)} \]
For independent events:
\[ P(J \cap K) = P(J) \cdot P(K) \]
Thus, we can rewrite:
\[ P(J|K) = \frac{P(J) \cdot P(K)}{P(K)} \]
Since \( P(K) \) is not zero, it cancels out from the numerator and denominator:
\[ P(J|K) = P(J) \]
Therefore, given:
\[ P(J|K) = 0.3 \]
We can conclude:
\[ P(J) = 0.3 \]
Hence, the probability \( P(J) \) is 0.3.
In summary, this problem illustrates the concept of independent events and conditional probability, demonstrating that for independent events, the conditional probability \( P(J|K) \) simply equals the probability \( P(J) \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faed2933e-b67f-42a0-9fac-b6dce0e828b0%2F98e3aba8-5a0d-49ae-bede-770b0ff15937%2Fua3quld_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding Independent Events in Probability
**Problem Statement:**
J and K are independent events. \( P(J|K) = 0.3 \). Find \( P(J) \).
**Solution:**
To solve this, we need to understand the basic concepts of independent events in probability.
**Key Concept:**
For two events J and K to be independent, the occurrence of one event does not affect the probability of the occurrence of the other event.
Mathematically, two events J and K are independent if and only if:
\[ P(J \cap K) = P(J) \cdot P(K) \]
**Given:**
\[ P(J|K) = 0.3 \]
By the definition of conditional probability:
\[ P(J|K) = \frac{P(J \cap K)}{P(K)} \]
For independent events:
\[ P(J \cap K) = P(J) \cdot P(K) \]
Thus, we can rewrite:
\[ P(J|K) = \frac{P(J) \cdot P(K)}{P(K)} \]
Since \( P(K) \) is not zero, it cancels out from the numerator and denominator:
\[ P(J|K) = P(J) \]
Therefore, given:
\[ P(J|K) = 0.3 \]
We can conclude:
\[ P(J) = 0.3 \]
Hence, the probability \( P(J) \) is 0.3.
In summary, this problem illustrates the concept of independent events and conditional probability, demonstrating that for independent events, the conditional probability \( P(J|K) \) simply equals the probability \( P(J) \).
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