In a class of $100$ students, $60$ students drink tea, $50$ students drink coffee and $30$ students drink both. A student from this class is selected at random, find the probability that student takes at least one of the two drinks.

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Solution

Let $T$ and $C$ be the number of students who drink tea and coffee respectively.
Then,
$P (T) = \frac{60}{100} = 0.6$
$P (C) = \frac{50}{100} = 0.5$
$P (T \cap C) = \frac{30}{100} = 0.3$
Therefore,
$P (T \cup C) = P (T) + P (C) - P (T \cap C)$
$P (T \cup C) = 0.6 + 0.5 - 0.3 = 0.8$

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In a class of 100 students, 60 students drink tea, 50 students drink coffee and 30 students drink both. A student from this class is selected at random, find the probability that student takes at least one of the two drinks.

Let T and C be the number of students who drink tea and coffee respectively. Then, P(T)=60100=0.6P(C)=50100=0.5P(T∩C)=30100=0.3Therefore, P(T∪C)=P(T)+P(C)−P(T∩C)P(T∪C)=0.6+0.5−0.3=0.8

In a class of $100$ students, $60$ students drink tea, $50$ students drink coffee and $30$ students drink both. A student from this class is selected at random, find the probability that student takes at least one of the two drinks.

Let $T$ and $C$ be the number of students who drink tea and coffee respectively.
Then,
$P (T) = \frac{60}{100} = 0.6$
$P (C) = \frac{50}{100} = 0.5$
$P (T \cap C) = \frac{30}{100} = 0.3$
Therefore,
$P (T \cup C) = P (T) + P (C) - P (T \cap C)$
$P (T \cup C) = 0.6 + 0.5 - 0.3 = 0.8$