NCERT Class 12 Maths: Probability Exercise 13.1 – Conditional Probability

Understanding Conditional Probability

Conditional probability is a fundamental concept in probability theory that deals with situations where the likelihood of an event occurring is dependent on the occurrence of another event. Imagine you're drawing cards from a deck. The probability of drawing a 'King' changes if you already know that the first card drawn (and not replaced) was an 'Ace'. This change in probability, based on prior information, is precisely what conditional probability addresses.

The conditional probability of an event E occurring, given that another event F has already occurred, is denoted as P(E|F). This is read as "the probability of E given F". The formula for conditional probability is given by:

\[P(E|F) = \frac{P(E \cap F)}{P(F)}\]

provided that P(F) > 0. Let's break down what each term means:

• P(E \cap F): This represents the probability of both events E and F occurring simultaneously. It's the probability of their intersection.
• P(F): This is the probability of event F occurring. It acts as our new (reduced) sample space in the context of event E.

The crucial insight here is the concept of a reduced sample space. When we are given that event F has already occurred, our universe of possible outcomes shrinks from the original sample space (S) to just the outcomes within event F. Then, we are interested in the outcomes within this new sample space F that also satisfy event E. Therefore, we are essentially looking for the proportion of outcomes in F that are also in E. The condition P(F) > 0 is essential because if P(F) were zero, it would mean event F is impossible, and conditioning on an impossible event is undefined.

Understanding this concept is key to solving problems in Exercise 13.1. Always start by clearly defining your events E and F, then calculate their individual probabilities and the probability of their intersection.

Key Definitions

Sample Space (S)

The set of all possible outcomes of a random experiment. For example, when rolling a single die, S = {1, 2, 3, 4, 5, 6}.

Event (E, F)

A subset of the sample space. It's a collection of one or more outcomes. For example, rolling an even number on a die is an event E = {2, 4, 6}.

Intersection of Events (E \cap F)

The event containing all outcomes that are common to both events E and F. It occurs when both E and F happen simultaneously.

Conditional Probability P(E|F)

The probability of event E occurring, given that event F has already occurred. It implies a reduction of the original sample space to event F.

Solved Examples from NCERT Ex 13.1

• Example 1: Basic Application Given P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find P(A \cap B). Step 1: Understand the given information. We are given the probabilities of two events A and B, and the conditional probability of B given A. P(A) = 0.8 P(B) = 0.5 P(B|A) = 0.4 We need to find P(A \cap B). Step 2: Recall the formula for conditional probability. The formula for P(B|A) is: P(B|A) = P(A \cap B) / P(A). Step 3: Substitute the known values into the formula and solve. 0.4 = P(A \cap B) / 0.8 P(A \cap B) = 0.4 0.8 P(A \cap B) = 0.32 Final Answer: P(A \cap B) = 0.32.
• Example 2: Dice Roll A die is thrown. Let E be the event "the number appearing is a multiple of 3" and F be the event "the number appearing is even". Find P(E|F). Step 1: Define the sample space and events. The sample space S for throwing a die is S = {1, 2, 3, 4, 5, 6}. Event E: "the number appearing is a multiple of 3". So, E = {3, 6}. Event F: "the number appearing is even". So, F = {2, 4, 6}. Step 2: Calculate individual probabilities P(E), P(F), and P(E \cap F). P(E) = (Number of outcomes in E) / (Total outcomes in S) = 2/6 = 1/3. P(F) = (Number of outcomes in F) / (Total outcomes in S) = 3/6 = 1/2. The intersection E \cap F contains outcomes common to E and F. E \cap F = {6}. P(E \cap F) = (Number of outcomes in E \cap F) / (Total outcomes in S) = 1/6. Step 3: Apply the conditional probability formula. P(E|F) = P(E \cap F) / P(F) P(E|F) = (1/6) / (1/2) P(E|F) = 1/6 * 2/1 P(E|F) = 2/6 = 1/3 Final Answer: The probability of getting a multiple of 3 given that an even number has appeared is 1/3.

Exam Tips for Conditional Probability

When solving conditional probability problems in your exams, always follow a systematic approach to avoid common errors:

1. Clearly Identify Events: Read the question carefully to determine which event is E (the event whose probability you want to find) and which event is F (the event that has already occurred or is given). Misidentifying E and F is a very common mistake.
2. Define Sample Space (S): List all possible outcomes of the experiment if not explicitly given. This helps in correctly calculating probabilities.
3. Calculate Individual Probabilities: Find P(E), P(F), and P(E \cap F) correctly. Remember that P(E \cap F) includes only those outcomes that are present in both E and F.
4. Check the Condition P(F) > 0: While rarely an issue in direct textbook problems, always remember that the conditioning event F must have a non-zero probability for P(E|F) to be defined.
5. Use the Formula Accurately: Substitute the calculated probabilities into the formula P(E|F) = P(E \cap F) / P(F) carefully. Double-check your arithmetic.
6. Don't Confuse P(E|F) with P(F|E): These are generally not the same. Make sure you're calculating the specific conditional probability asked for.

Practice Questions with Solutions

• Q: A pair of dice is thrown. Let E be the event that 'the sum of the numbers on the dice is 8' and F be the event that 'a 3 appears on at least one die'. Find P(E|F). A: Step 1: Define the sample space and events. The total sample space S for throwing two dice has 6 * 6 = 36 outcomes. Event E (sum is 8): E = {(2,6), (3,5), (4,4), (5,3), (6,2)}. So, n(E) = 5. Event F (3 appears on at least one die): F = {(1,3), (2,3), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,3), (5,3), (6,3)}. So, n(F) = 11. Step 2: Find the intersection E \cap F. E \cap F = {(3,5), (5,3)}. So, n(E \cap F) = 2. Step 3: Calculate probabilities. P(F) = n(F) / n(S) = 11/36. P(E \cap F) = n(E \cap F) / n(S) = 2/36. Step 4: Apply the conditional probability formula. P(E|F) = P(E \cap F) / P(F) = (2/36) / (11/36) = 2/11. Final answer: P(E|F) = 2/11.
• Q: In a school, 30% of students prefer Maths, 20% prefer Science, and 10% prefer both. If a student is chosen randomly, and it is known that they prefer Maths, what is the probability that they also prefer Science? A: Step 1: Define events and given probabilities. Let M be the event that a student prefers Maths. P(M) = 0.30. Let S be the event that a student prefers Science. P(S) = 0.20. Let M \cap S be the event that a student prefers both. P(M \cap S) = 0.10. We need to find P(S|M), the probability that a student prefers Science given they prefer Maths. Step 2: Apply the conditional probability formula. P(S|M) = P(S \cap M) / P(M). Step 3: Substitute values and calculate. P(S|M) = 0.10 / 0.30 = 1/3. Final answer: P(S|M) = 1/3.
• Q: A card is drawn from a well-shuffled deck of 52 cards. What is the probability that the card is a 'King', given that it is a 'face card'? A: Step 1: Define sample space and events. Total cards in a deck = 52. Let K be the event that the card drawn is a King. There are 4 Kings (K_heart, K_diamond, K_club, K_spade). So, n(K) = 4. Let F be the event that the card drawn is a face card. There are 12 face cards (4 Kings, 4 Queens, 4 Jacks). So, n(F) = 12. Step 2: Find the intersection K \cap F. The intersection K \cap F is the event that the card is both a King and a face card. All Kings are face cards, so K \cap F = K. Thus, n(K \cap F) = 4. Step 3: Calculate probabilities. P(F) = n(F) / 52 = 12/52. P(K \cap F) = n(K \cap F) / 52 = 4/52. Step 4: Apply the conditional probability formula. P(K|F) = P(K \cap F) / P(F) = (4/52) / (12/52) = 4/12 = 1/3. Final answer: P(K|F) = 1/3.
• Q: If P(E) = 0.6, P(F) = 0.3 and P(E \cap F) = 0.2, find P(E|F) and P(F|E). A: Step 1: Given probabilities. P(E) = 0.6 P(F) = 0.3 P(E \cap F) = 0.2 Step 2: Calculate P(E|F). P(E|F) = P(E \cap F) / P(F) = 0.2 / 0.3 = 2/3. Step 3: Calculate P(F|E). P(F|E) = P(F \cap E) / P(E). Note that P(F \cap E) is the same as P(E \cap F). P(F|E) = 0.2 / 0.6 = 2/6 = 1/3. Final answer: P(E|F) = 2/3 and P(F|E) = 1/3.

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