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GATE CSEIT/Engineering Mathematics/ Conditional probability ( With short trick).

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GATE CSEIT/Engineering Mathematics/ Conditional probability ( With short trick).

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​Conditional probability is a fundamental concept in probability theory, especially relevant for the GATE Computer Science and Information Technology (CSEIT) exam. It measures the likelihood of an event occurring given that another event has already occurred.​

Definition:

If AA and BB are two events, the conditional probability of AA given BB is defined as: P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)} provided P(B)≠0P(B) \neq 0.​

Shortcut Techniques for GATE Problems:

  1. Utilize Symmetry: In problems where outcomes are symmetrical, symmetry can simplify calculations. For example, when dealing with fair coins or unbiased dice, each outcome has an equal probability, allowing for straightforward computations.

  2. Complementary Counting: Sometimes, it’s easier to calculate the probability of the complement of an event and subtract from 1: P(A)=1−P(A′)P(A) = 1 – P(A’) This approach is particularly useful when the event’s complement has a simpler or more direct calculation.

  3. Bayes’ Theorem: For problems involving reverse conditional probabilities, Bayes’ Theorem is invaluable: P(A∣B)=P(B∣A)⋅P(A)P(B)P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} This theorem allows the computation of P(A∣B)P(A|B) using P(B∣A)P(B|A), which might be more straightforward to determine.

  4. Total Probability Theorem: When an event can occur through multiple mutually exclusive scenarios, the total probability is the sum of the probabilities of each scenario: P(B)=∑iP(B∣Ai)⋅P(Ai)P(B) = \sum_{i} P(B|A_i) \cdot P(A_i) This technique is useful for breaking down complex problems into manageable parts.

Practice Resources:

Example Problem:

​An urn contains 5 red balls and 5 black balls. A ball is drawn at random and discarded without observing its color. Then, another ball is drawn. What is the probability that the second ball is red?​

Solution:

Since the first ball’s color is unknown and discarded, the total number of balls remains 9 for the second draw. The probability of drawing a red ball in the second draw is: P(Red)=Number of red ballsTotal number of remaining balls=59P(\text{Red}) = \frac{\text{Number of red balls}}{\text{Total number of remaining balls}} = \frac{5}{9}​

Understanding and applying these shortcut techniques can enhance problem-solving efficiency in the GATE exam.

GATE CSEIT/Engineering Mathematics/ Conditional probability ( With short trick).

Conditional Probability

Here’s a focused and simplified guide on Conditional Probability for GATE CSE/IT – Engineering Mathematics, including a shortcut trick and examples.

What is Conditional Probability?

Conditional probability is the probability of an event A occurring given that another event B has already occurred.

Formula:

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

Where:

Shortcut Trick (GATE Focused)

“Restrict the Sample Space to B”
Instead of thinking in terms of full probability, narrow your focus only to cases where B happens, and from those, find how many also satisfy A.

Trick in Action:

Suppose:
A box contains 5 red balls and 3 blue balls. One ball is drawn at random, and it’s known that the ball is not blue. What is the probability that it is red?

Solution using trick:

  1. Total favorable cases (non-blue): 5 (only red)

  2. Required outcome (red | not blue): 5/5 = 1

Because the condition “not blue” restricts the sample space to red balls only, so the probability of red given not blue is 1.

Common GATE-Type Example

Q: A card is drawn from a well-shuffled deck. What is the probability that it is an ace given that it is a red card?

Given:

P(Ace∣Red)=P(Ace∩Red)P(Red)=226=113P(Ace | Red) = \frac{P(Ace \cap Red)}{P(Red)} = \frac{2}{26} = \frac{1}{13}

Practice Problem (GATE Pattern)

Q: Two fair dice are thrown. What is the probability that the sum is 8, given that one die shows a 3?

Step-by-step using shortcut:

  1. Total outcomes where one die is 3:

    • (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

    • (1,3), (2,3), (4,3), (5,3), (6,3)
      => Total = 11 outcomes

  2. Outcomes among these where sum = 8:

    • (3,5), (5,3)

    => Favorable = 2

P(Sum=8∣Onedieis3)=211P(Sum = 8 \mid One die is 3) = \frac{2}{11}

Answer: 211\frac{2}{11}

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GATE CSEIT/Engineering Mathematics/ Conditional probability ( With short trick).

Theory of Probability – Engineering Mathematics