A) zero
B) \[\frac{1}{3}\]
C) \[\frac{1}{4}\]
D) None of these
Correct Answer: D
Solution :
Given numbers are 1, 2, 3, 4, Possibilities for unit?s place digit (either 1 or 3) = 2 Possibilities for ten?s digit = 3 Possibilities for thousand? place?s digit = 2 \[\therefore \] Number of favourable outcomes \[=2\times 3\times 2\times 1=12\] Number of numbers formed by 1, 2, 3, 4 (without repetitions ) = 4! \[\therefore \] Required probability \[=\frac{12}{4\times 3\times 2}=\frac{1}{2}\]
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