MathematicsA bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag. Find the probability of getting neither a green ball nor a red ball.

A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag. Find the probability of getting neither a green ball nor a red ball.


  1. A
    1720
  2. B
    720
  3. C
    120
  4. D
    320 

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    Solution:

    Given that there are 5 red, 7 white and 8 green balls in a bag.
    We know that the probability is given as the ratio of the number of favorable outcomes with the total number of possible outcomes.
    P(E)=Number of favourable outcomes n(E)Total possible outcomes n(S)
    Total number of balls are:
    5+7+8=20
    n(S)=20
    Let E be the event of getting neither a green nor a red ball.
    The probability of getting neither a green nor a red ball is equal to the probability of getting a white ball.
    Total number of white balls is 7.
    n(E)=7
    So the required probability is,
    P(E)=Number of favourable outcomes n(E)Total number of outcomes n(S)
    P(E)=720
    Thus, the probability of getting neither green nor a red ball is 720.
    Hence, option 2 is correct.
     
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