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14.Probability
hard
The probability that a randomly chosen $5-digit$ number is made from exactly two digits is
A
$\frac{121}{10^{4}}$
B
$\frac{150}{10^{4}}$
C
$\frac{135}{10^{4}}$
D
$\frac{134}{10^{4}}$
(JEE MAIN-2020)
Solution
First Case: Choose two non-zero digits ${ }^{9} C _{2}$
Now, number of 5 -digit numbers containing both digits $=2^{5}-2$
Second Case: Choose one non-zero \& one zero as digit ${ }^{9} C _{1}$
Number of 5 -digit numbers containg one non zero and one zero both $=\left(2^{4}-1\right)$ Required prob.
$=\frac{\left({ }^{9} C _{2} \times\left(2^{5}-2\right)+{ }^{9} C _{1} \times\left(2^{4}-1\right)\right)}{9 \times 10^{4}}$
$=\frac{36 \times(32-2)+9 \times(16-1)}{9 \times 10^{4}}$
$=\frac{4 \times 30+15}{10^{4}}=\frac{135}{10^{4}}$
Standard 11
Mathematics
