If a digit is chosen at random from the digits 1, 2, 3, 4, 5, 6, 7, 8,

1.	If a digit is chosen at random from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, then the probability that it is odd, isA \(\frac{4}9\)B \(\frac{5}9\)C \(\frac{1}9\)D \(\frac{2}3\)
Answer» Total numbers of elementary events are = 9 Let E be the event of getting an odd number The favorable outcomes are: 1, 3, 5, 7, 9 Numbers of favorable events= 5 P (odd number) = P (E) = \(\frac{5}9\)

If a digit is chosen at random from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, then the probability that it is odd, isA \(\frac{4}9\)B \(\frac{5}9\)C \(\frac{1}9\)D \(\frac{2}3\)

Answer»

Total numbers of elementary events are = 9

Let E be the event of getting an odd number

The favorable outcomes are: 1, 3, 5, 7, 9

Numbers of favorable events= 5

P (odd number) = P (E) = \(\frac{5}9\)

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If a digit is chosen at random from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, then the probability that it is odd, isA \(\frac{4}9\)B \(\frac{5}9\)C \(\frac{1}9\)D \(\frac{2}3\)

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