Sounds like a homework/test question.
We can find an exact answer to this question. There are (52c5) = 2598960 total combinations of five card Poker hands. We are interested in the ones where we are dealt two pair.
Let's count all the hands that have two pair and no more than two pair (quads or full house). Each pair has (4c2) = 6 forms. Of the 13 ranks there (13c2) = 78 ways to select the two pair. Once both pairs are selected there are 44 cards left for the remaining card. So there are 78 *6 *6 * 44 = 123552 hands with two pair.
In decimal form simply for 123552 / 2598960 = 4.7539015606242497614e-02 or about a 4.8% chance.
Probability that it is one of these eight cards is 8/52. Hence the probability of not getting these eight cards is 44/52
Three over eight as a decimal is 0.375
8.8 IS a decimal. Eight and eight-tenths 8 8/10
Seven and eight tenths in a decimal is 7.8.
800 and 8/100 = 800.08 as a decimal
A 8/52 chance.
Eight and eight hundredths in a decimal is 8.08
eight and eight hundredths = 8.08 in decimaleight and eight hundredths as a decimal = 8.08
The probability of rolling a four on an eight sided octahedron is 1 in 8, or 0.125.
Fifty-eight thousands in decimal form is 58,000Fifty-eight thousandths in decimal form is 0.058
Probability that it is one of these eight cards is 8/52. Hence the probability of not getting these eight cards is 44/52
eleven and eight hundredths in decimal = 11.08
eight tenths as a decimal = 0.8
Three over eight as a decimal is 0.375
one over eight to a decimal = 0.125
eighteen and eight hundredths = 18.08 in decimal
Eight millionths in decimal numbers is 0.000008