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An urn contains $9$ red, $7$ white and $4$ black balls A ball is drawn at random. What is the probability that the ball drawn is not red?
$
  A.{\text{ }}\dfrac{1}{{11}} \\
  B.{\text{ }}\dfrac{9}{{11}} \\
  C.\,\;\dfrac{2}{{11}} \\
  D.{\text{ }}\dfrac{{11}}{{20}} \\
 $

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Last updated date: 13th Jun 2024
Total views: 412.8k
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Answer
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Hint: Probability is the state of being probable and the extent to which something is likely to happen in the particular situations or the favourable outcomes. Probability of any given event is equal to the ratio of the favourable outcomes with the total number of the outcomes.
$P(A) = $ Total number of the favourable outcomes / Total number of the outcomes

Complete step by step solution: Here, the total number of possible outcomes is equal to the sum of total number of all the balls in an urn.
Total = Sum of red, white and the black balls.
$
  n(S) = 9 + 7 + 4 \\
  n(S) = 20 \\
 $
Now, let us suppose that A be an event that the drawn ball is not red.
(So, we will count all the possibility of the balls excluding the red coloured balls)
Therefore, the favourable outcomes will be the summation of white balls and black balls.
$
  \therefore n(A) = 7 + 4 \\
  \therefore n(A) = 11 \\
 $
Therefore the required probability that the drawn ball is not red is –
$
  P(A) = \dfrac{{n(A)}}{{n(S)}} \\
  P(A) = \dfrac{{11}}{{20}} \\
 $
Therefore, the probability that the ball drawn is not red is $\dfrac{{11}}{{20}}$
Hence, from the given multiple choices, option D is the correct answer.

Note: The probability of any event always ranges between zero and one. It can never be the negative number or the number greater than one. The probability of impossible events is always equal to zero whereas, the probability of the sure event is always equal to one.