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, is
A \(\frac{4}9\)
B \(\frac{5}9\)
C \(\frac{1}9\)
D \(\frac{2}3\)
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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, is
A \(\frac{4}9\)
B \(\frac{5}9\)
C \(\frac{1}9\)
D \(\frac{2}3\)
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, is
A \(\frac{4}9\)
B \(\frac{5}9\)
C \(\frac{1}9\)
D \(\frac{2}3\)
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) = \(\frac{5}9\)