A five digit number is selected at random. The probability that the digits in the odd places are odd and the even places are even. (Repetition is not allowed).
a)5P2×5P3÷10^4×9
b)5P2×5P3÷10^5
c)5C2×5C3×2÷10^4×9
d)5C2×5C3÷10^4×9
Answers
Given : A five digit number is selected at random.
To Find : The probability that the digits in the odd places are odd and the even places are even. (Repetition is not allowed).
Solution:
total odd positions = 3
total odd digits = 5
ways of selection and arranging = 5P3
total even positions = 2
total even digits = 5
ways of selection and arranging = 5P2
total possible numbers
1st digit can not be zero hence
9 ways remaining 4 out of 9 in 9P4
ways
probability = 5P2 x 5P3 / (9 × 9P4)
none of the given option matches.
if assuming repetition allowed in total numbers then
9 × 10^4 total numbers
probability = 5P2 × 5P3 / 9 × 10^4
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