Question

There are six cards in a box, each bearing a number from 0 to 5. Find the probability of each of the following events, that a card drawn shows,
(1) a natural number.
(2) a number less than 1.
(3) a whole number.
(4) a number is greater than 5.
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Answer 1

Total number of cards = 6
(1)
Number of cards bearing a natural number = 1, 2, 3, 4, 5
= 5
Therefore, required probability = $\frac{\mathrm{Number}\mathrm{of}\mathrm{cards}\mathrm{bearing}\mathrm{a}\mathrm{natural}number}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{cards}}$
$=\frac{5}{6}$
Hence, the probability that a card drawn shows a natural number is $\frac{5}{6}$.
(2)
Number of cards bearing a number less than 1 = 1
Therefore, required probability = $\frac{\mathrm{Number}\mathrm{of}\mathrm{cards}\mathrm{bearing}\mathrm{a}number\mathrm{less}\mathrm{than}1}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{cards}}$
$=\frac{1}{6}$
Hence, the probability that a card drawn shows a number less than 1 is $\frac{1}{6}$.
(3)
Number of cards bearing a whole number = 0, 1, 2, 3, 4, 5
= 6
Therefore, required probability = $\frac{\mathrm{Number}\mathrm{of}\mathrm{cards}\mathrm{bearing}\mathrm{a}\mathrm{whole}number}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{cards}}$
$=\frac{6}{6}$
= 1
Hence, the probability that a card drawn shows a whole number is 1.
(4)
Number of cards bearing a number greater than 5 = 0
Therefore, required probability = $\frac{\mathrm{Number}\mathrm{of}\mathrm{cards}\mathrm{bearing}\mathrm{a}\mathrm{number}\mathrm{greater}\mathrm{than}5}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{cards}}$
$=\frac{0}{6}$
= 0
Hence, the probability that a card drawn shows a number greater than 5 is 0.