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NIELIT Scientist B 2020 November: 38

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In a group of 24 members, each member drinks either tea or coffee or both. If $15$ of them drink tea and $18$ drink coffee, find the probability that a person selected from the group drinks both tea and coffee.

  1. $1/8$
  2. $3/8$
  3. $5/24$
  4. None of the options
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Given that, $n(T) = 15, n(C) = 18,$ and $n(T\cup C) = 24$

Now, $n(T\cup C) = n(T) + n(C) – n(T \cap C)$

$\implies n(T \cap C) = 15+18-24 = 9$

Required probability $ = \dfrac{9}{24} = \dfrac{3}{8}.$

So, the correct answer is $(B).$
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