If A and B are two events such that P(not A) = \(\rm \frac{7}{10}\) , P(not B) = \(\rm \frac{3}{10}\) and P(A|B) = \(\rm \frac{3}{14}\) , then what is P(B|A) equal to?
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
| Quantity | Calculation | Result |
|---|---|---|
| P(A) | \(1 - P(\text{not A})\) | \(\frac{3}{10}\) |
| P(B) | \(1 - P(\text{not B})\) | \(\frac{7}{10}\) |
| P(A \(\cap\) B) | \(P(A|B) \times P(B)\) | \(\frac{3}{14} \times \frac{7}{10} = \frac{3}{20}\) |
| P(B|A) | \(\frac{P(A \cap B)}{P(A)}\) | \(\frac{\frac{3}{20}}{\frac{3}{10}} = \frac{1}{2}\) |
Revision Table: Key Probability Concepts
| Concept | Formula | Description |
|---|---|---|
| Probability of Complement | P(E) = 1 - P(not E) | The probability of an event is 1 minus the probability of its complement. |
| Conditional Probability P(A|B) | \(P(A|B) = \frac{P(A \cap B)}{P(B)}\) | Probability of event A given that event B has occurred. P(B) > 0. |
| Conditional Probability P(B|A) | \(P(B|A) = \frac{P(A \cap B)}{P(A)}\) | Probability of event B given that event A has occurred. P(A) > 0. |
| Probability of Intersection | \(P(A \cap B) = P(A|B) \times P(B)\) | Probability of both A and B occurring. Also \(P(A \cap B) = P(B|A) \times P(A)\). |
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.
Similar Questions
What is \(P(\overline T | \overline G)\) equal to?
What is \(P(G | \overline T) \) equal to?
What is \(P (G \cap \overline T)\) equal to?
Let two events A and B be such that P(A) = L and P(B) = M. Which one of the following is correct?
If P(A|B) < P(A), then which one of the following is correct?
A and B are two events such that A̅ and B̅ are mutually exclusive. If P(A) = 0.5 and P(B) = 0.6, then what is the value of P(A|B)?
Three groups of children contain 3 girls and 1 boy; 2 girls and 2 boys: 1 girl and 3 boys. One child is selected at random from each group. The probability that the three selected consist of 1 girl and 2 boys is
Two integers x and y are chosen with replacement from the set (0, 1, 2…10). The probability that |x - y| > 5 is
If two dice are thrown and at least one the dice show 5, then the probability that the sum is 10 or more is
For two dependent events A and B, it is given that P(A) = 0.2 and P(B) = 0.5. If A ⊆ B, then the values of conditional probabilities P(A|B) and P(B|A) are respectively
Important Questions from Conditional Probability
If P(A) = 0.7, P(B) = 0.5 and P(B/A) = 0.3, find (i) P(A/B) (ii) P(A ∪ B)?
Let A and B be two events such that \(P(\overline{A \cup B}) = \dfrac{1}{6}\) , P(A ∩ B) = \(\dfrac{1}{4}\) and P(A̅) = \(\dfrac{1}{4}\) , where A̅ stands for complement of event A. Then, events A and B are:
A bike manufacturing factory has two plants P and Q. Plant P manufactures 60 percent of bikes and plant Q manufacture 40 percent. 80 percent of the bikes at plant P and 90 percent of the bikes at plant Q are rated of standard quality. A bike is chosen at random and is found to be of standard quality. What is the probability that it has come from plant P?
20 percent of the pens produced in a factory are of red colour and 4 percent are red and defective. If one pen is picked up at random, then what is the probability of its being defective if it is red?
In a game, there are three rooms- I, Il and IIl. Room I contain 2 boxes having gift items and 3 empty boxes, room II contains 3 boxes having gift items and 2 empty boxes, and room III contains 4 boxes having gift items and one empty box respectively. There is an equal probability of each room being chosen by a player. Mr John selects one box from a room chosen at random. The probability that Mr John wins a box having gift items is:
