12 thoughts on “Probability – 18

  1. Thank you sir for such enlightening post.

    Sir i would like to point out that there is no difference between the first and the second question. The framing of the first question looks wrong. It should have been first one an ace, second one a king and third one a queen. Please correct me if i am wrong.

  2. Thank you for the post.

    In the 3rd question why do we divide by 2!? I thought the aces selected are distinct and thus can be arranged in 3! ways.

    • Because we have counted AHeart, ASpade, KHeart as a separate case from ASpade, AHeart, KHeart, while in fact they are a single case (since we are not arranging the cards once picked)

      regards
      J

  3. In the last question the answer (4c2*4c1)/52C3 i feel confused. If you are using 4C2 then how will you deal with it if the cards are taken simultaneously.As per my knowledge, the answer is (4C1*3C1*4C1)/52C3

  4. Sir,help needed !

    A bag contains 10 balls numbered from 0 to 9. The balls are such that the person picking a ball out of the bag is equally likely to pick any one of them. A person picked a ball and replaced it in the bag after noting its number. He repeated this process 2 more times. What is the probability that the ball picked first is numbered higher than the ball picked second, and the ball picked second is numbered higher than the ball picked third?

    i am stuck with the permutations for replacements

    9/10 x 8/10 x 7/10

    what will be the correct approach?

      • Thanks a ton sir 😀 i was going totally mad over this. Sir,can you suggest me anything,to strong my ratios and proportion verticle, to tackle questions with change in volume?

  5. At this time, don’t try to tackle a new concept or topic. A little knowledge is a dangerous thing etc etc – chances are you will end up messing up because of insufficient practice…stick to your strengths and make them impregnable

    regards
    J

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