{"id":3362,"date":"2025-06-07T04:39:19","date_gmt":"2025-06-07T04:39:19","guid":{"rendered":"https:\/\/diznr.com\/?p=3362"},"modified":"2025-06-07T04:39:19","modified_gmt":"2025-06-07T04:39:19","slug":"gate-cseit-engineering-mathematics-conditional-probability-with-trick-short","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/pdf\/gate-cseit-engineering-mathematics-conditional-probability-with-trick-short\/","title":{"rendered":"GATE CSEIT\/Engineering Mathematics\/ Conditional probability ( With short trick)."},"content":{"rendered":"

GATE CSEIT\/Engineering Mathematics\/ Conditional probability ( With short trick).<\/p>\n

[fvplayer id=”371″]<\/p>\n

\u200bConditional probability is a fundamental concept in probability theory, especially relevant for the GATE Computer Science and Information Technology (CSEIT) exam.<\/span> It measures the likelihood of an event occurring given that another event has already occurred.<\/span>\u200b<\/p>\n

Definition:<\/strong><\/p>\n

If AA<\/span>A<\/span><\/span><\/span><\/span> and BB<\/span>B<\/span><\/span><\/span><\/span> are two events, the conditional probability of AA<\/span>A<\/span><\/span><\/span><\/span> given BB<\/span>B<\/span><\/span><\/span><\/span> is defined as:<\/span> P(A\u2223B)=P(A\u2229B)P(B)P(A|B) = \\frac{P(A \\cap B)}{P(B)}<\/span>P<\/span>(<\/span>A<\/span>\u2223<\/span>B<\/span>)<\/span>=<\/span><\/span>P<\/span>(<\/span>B<\/span>)<\/span><\/span><\/span>P<\/span>(<\/span>A<\/span>\u2229<\/span>B<\/span>)<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> provided P(B)\u22600P(B) \\neq 0<\/span>P<\/span>(<\/span>B<\/span>)<\/span>\ue020<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>0<\/span><\/span><\/span><\/span>.<\/span>\u200b<\/p>\n

Shortcut Techniques for GATE Problems:<\/strong><\/p>\n

    \n
  1. \n

    Utilize Symmetry:<\/strong> 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.<\/p>\n<\/li>\n

  2. \n

    Complementary Counting:<\/strong> Sometimes, it’s easier to calculate the probability of the complement of an event and subtract from 1: P(A)=1\u2212P(A\u2032)P(A) = 1 – P(A’)<\/span>P<\/span>(<\/span>A<\/span>)<\/span>=<\/span><\/span>1<\/span>\u2212<\/span><\/span>P<\/span>(<\/span>A<\/span>\u2032<\/span><\/span><\/span><\/span><\/span><\/span><\/span>)<\/span><\/span><\/span><\/span> This approach is particularly useful when the event’s complement has a simpler or more direct calculation.<\/p>\n<\/li>\n

  3. \n

    Bayes’ Theorem:<\/strong> For problems involving reverse conditional probabilities, Bayes’ Theorem is invaluable: P(A\u2223B)=P(B\u2223A)\u22c5P(A)P(B)P(A|B) = \\frac{P(B|A) \\cdot P(A)}{P(B)}<\/span>P<\/span>(<\/span>A<\/span>\u2223<\/span>B<\/span>)<\/span>=<\/span><\/span>P<\/span>(<\/span>B<\/span>)<\/span><\/span><\/span>P<\/span>(<\/span>B<\/span>\u2223A<\/span>)<\/span>\u22c5<\/span>P<\/span>(<\/span>A<\/span>)<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> This theorem allows the computation of P(A\u2223B)P(A|B)<\/span>P<\/span>(<\/span>A<\/span>\u2223<\/span>B<\/span>)<\/span><\/span><\/span><\/span> using P(B\u2223A)P(B|A)<\/span>P<\/span>(<\/span>B<\/span>\u2223<\/span>A<\/span>)<\/span><\/span><\/span><\/span>, which might be more straightforward to determine.<\/p>\n<\/li>\n

  4. \n

    Total Probability Theorem:<\/strong> When an event can occur through multiple mutually exclusive scenarios, the total probability is the sum of the probabilities of each scenario: P(B)=\u2211iP(B\u2223Ai)\u22c5P(Ai)P(B) = \\sum_{i} P(B|A_i) \\cdot P(A_i)<\/span>P<\/span>(<\/span>B<\/span>)<\/span>=<\/span><\/span>\u2211<\/span>i<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>P<\/span>(<\/span>B<\/span>\u2223<\/span>A<\/span>i<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>)<\/span>\u22c5<\/span><\/span>P<\/span>(<\/span>A<\/span>i<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>)<\/span><\/span><\/span><\/span> This technique is useful for breaking down complex problems into manageable parts.<\/p>\n<\/li>\n<\/ol>\n

    Practice Resources:<\/strong><\/p>\n