Understanding the Probability Problem

The problem asks us to find the conditional probability of event B occurring given that event A has occurred, denoted as P(B|A). We are provided with the probabilities of the complements of events A and B, and the conditional probability of A given B, i.e., P(not A), P(not B), and P(A|B).

Given Information:

• P(not A) = \(\rm \frac{7}{10}\)
• P(not B) = \(\rm \frac{3}{10}\)
• P(A|B) = \(\rm \frac{3}{14}\)

Calculating Basic Probabilities P(A) and P(B)

We know that the probability of an event occurring plus the probability of its complement not occurring is equal to 1. That is, P(E) + P(not E) = 1. We can use this to find P(A) and P(B).

• Probability of A, P(A):
  P(A) = 1 - P(not A)
  P(A) = \(1 - \frac{7}{10} = \frac{10 - 7}{10} = \frac{3}{10}\)
• Probability of B, P(B):
  P(B) = 1 - P(not B)
  P(B) = \(1 - \frac{3}{10} = \frac{10 - 3}{10} = \frac{7}{10}\)

Finding the Probability of the Intersection P(A \(\cap\) B)

The formula for conditional probability P(A|B) is given by:

\(P(A|B) = \frac{P(A \cap B)}{P(B)}\)

We can rearrange this formula to find P(A \(\cap\) B):

\(P(A \cap B) = P(A|B) \times P(B)\)

Substitute the given value of P(A|B) and the calculated value of P(B):

\(P(A \cap B) = \frac{3}{14} \times \frac{7}{10}\)

Let's calculate the product:

\(P(A \cap B) = \frac{3 \times 7}{14 \times 10} = \frac{21}{140}\)

We can simplify the fraction:

\(P(A \cap B) = \frac{21 \div 7}{140 \div 7} = \frac{3}{20}\)

Calculating the Conditional Probability P(B|A)

The formula for the conditional probability P(B|A) is given by:

\(P(B|A) = \frac{P(A \cap B)}{P(A)}\)

Now, substitute the calculated values for P(A \(\cap\) B) and P(A):

\(P(B|A) = \frac{\frac{3}{20}}{\frac{3}{10}}\)

To divide fractions, we multiply the numerator by the reciprocal of the denominator:

\(P(B|A) = \frac{3}{20} \times \frac{10}{3}\)

Multiply the fractions:

\(P(B|A) = \frac{3 \times 10}{20 \times 3} = \frac{30}{60}\)

Simplify the result:

\(P(B|A) = \frac{30 \div 30}{60 \div 30} = \frac{1}{2}\)

Thus, the value of P(B|A) is \(\frac{1}{2}\).

Summary of Calculations

Revision Table: Key Probability Concepts

Additional Information: Bayes' Theorem Connection

This problem can also be linked to Bayes' Theorem. Bayes' Theorem provides a way to update the probability of a hypothesis based on new evidence. The basic form is:

\(P(B|A) = \frac{P(A|B) \times P(B)}{P(A)}\)

Notice that the formula we used, \(P(B|A) = \frac{P(A \cap B)}{P(A)}\), combined with the definition \(P(A \cap B) = P(A|B) \times P(B)\), directly leads to Bayes' Theorem. In this problem, we calculated P(A) and P(B) first, then P(A \(\cap\) B) using P(A|B) and P(B), and finally P(B|A) using P(A \(\cap\) B) and P(A). This step-by-step calculation effectively applies the underlying principles used in Bayes' Theorem.