Cards are marked from $10$ to $100 .$ A card is chosen at random, what is the probability of getting a multiple of $9 ?$

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Solution

The number that are multiple between $0$ to $100$ is $18, 27, 36, 45, 54, 63, 72, 81, 90, 99$
$n (A) = 10$
and
$n (T) = 90$
Now
$\begin{matrix} p r o b a b i l i t y o f g e t t i n g m u l t i p l e o f 9 = \frac{n (A)}{n (T)} = \frac{10}{90} = \frac{1}{9} w h i c h i s t h e r e q u i r e d p r o b a b i l i t y . \end{matrix}$

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Cards are marked from 10 to 100. A card is chosen at random, what is the probability of getting a multiple of 9?

The number that are multiple between 0 to 100 is 18,27,36,45,54,63,72,81,90,99n(A)=10andn(T)=90Nowprobabilityofgettingmultipleof9=n(A)n(T)=1090=19whichistherequiredprobability.

Cards are marked from $10$ to $100 .$ A card is chosen at random, what is the probability of getting a multiple of $9 ?$