what is the probability that a two digit number selected at random is a multiple of 3 and not a multiple of 5?	Choose one:  A) 1 / 15  B) 2 / 15  C) 4 / 15  D) 7 / 15  E) none of these
<gadget id="calculator">5 * 2</gadget>
<output>10</output>

<gadget id="calculator">10 * 3</gadget>
<output>30</output>

<gadget id="calculator">3 * 2</gadget>
<output>6</output>

<gadget id="calculator">30 - 6</gadget>
<output>24</output>

<gadget id="calculator">3 * 3</gadget>
<output>9</output>

<gadget id="calculator">10 * 9</gadget>
<output>90</output>

<gadget id="calculator">24 / 90</gadget>
<output>4/15 = around 0.266667</output>

<result>C</result>
what is the probability that a two digit number selected at random is a multiple of 3 and not a multiple of 5?
divide(subtract(multiply(multiply(5, const_2), 3), multiply(3, const_2)), multiply(multiply(5, const_2), multiply(3, 3)))
"explanation : since every third number starting from 10 will be divisible by 3 , so total number of numbers divisible by 3 are 90 / 3 = 30 numbers which are divisible by 3 and 5 both are numbers which are multiple of 15 . for the range 10 to 99 , 15 is the first number divisible by 15 and 90 is the last number . so total number of numbers divisible by 15 are : ( 90 - 15 ) / 15 + 1 = 5 + 1 = 6 number of numbers which are divisible by 3 are 30 and number of numbers which are divisible by 3 and 5 both are 6 . so number of numbers divisible by 3 and not by 5 are : 30 - 6 = 24 so total probability = 24 / 90 = 4 / 15 answer : c"