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14.Probability
normal
Out of all possible $8$ digit numbers formed using all the digits $0,0,1,1,2,3,4,4$ a number is randomly selected. Probability that the selected number is odd, is-
A
$\frac{5}{7}$
B
$\frac{5}{9}$
C
$\frac{5}{11}$
D
$\frac{5}{14}$
Solution
Total number of numbers $ = {\,^7}{{\rm{C}}_2} \cdot \frac{{6!}}{{2!2!}} = {\rm{a}}$
for odd numbers:
Numbers with unit's digit $3:$
$^{6} \mathrm{C}_{2} \cdot \frac{5 !}{2 ! 2 !}=\mathrm{b}$
Numbers with unit's digit $1:$
$^{6} \mathrm{C}_{2} \cdot \frac{5 !}{2 !}=\mathrm{c}$
$\therefore $ Required probability $=\frac{b+c}{a}=\frac{5}{14}$
Standard 11
Mathematics
