The probability that a randomly chosen 5-digi | vedclass

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Question

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 on

Vedclass · Accepted Answer

$\frac{135}{10^{4}}$

Vedclass · Answer

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 $=\left(2^{4}-1ight)$ Required prob.

$=\frac{\left({ }^{9} C _{2} 	imes\left(2^{5}-2ight)+{ }^{9} C _{1} 	imes\left(2^{4}-1ight)ight)}{9 	imes 10^{4}}$

$=\frac{36 	imes(32-2)+9 	imes(16-1)}{9 	imes 10^{4}}$

$=\frac{4 	imes 30+15}{10^{4}}=\frac{135}{10^{4}}$

The probability that a randomly chosen $5-digit$ number is made from exactly two digits is

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