<PRE><B>Question 693600</B>
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In the numbers 1 to 10 there are exactly 4 prime numbers, namely 2, 3, 5, and 7.


The probability that the first card has a prime number is *[tex \LARGE \frac{4}{10}\ =\ \frac{2}{5}]


The probability that the second card has a prime number, given that the first had a prime number is *[tex \LARGE \frac{3}{9}\ =\ \frac{1}{3}], since there are now only 3 prime number cards left out of 9 total cards left to choose from.


Since the two events described above are independent, the probability of both events is the product of the two probabilities.  Multiply *[tex \LARGE \frac{2}{5}] times *[tex \LARGE \frac{1}{3}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
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