Originally Posted by

**janvdl** I'm not really good with probability, so excuse me if these are not completely correct.

For 1)

We have 8 counters in total, the chances of drawing a green counter would be 5 out of 8, which is 62,5 %

When the 2nd counter is drawn, there are only 7 counters left.

Chances of drawing a red counter would then be 3 out of 7, which is a chance of 42,9%

The chances that the first one is green AND the second one is red is:

$\displaystyle \frac{5}{8} \times \frac{3}{7}$

$\displaystyle = \frac{15}{56}$

$\displaystyle = 26,8$ %

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For 2)

The chances of drawing two green counters are:

$\displaystyle \frac{5}{8} \times \frac{4}{7}$

$\displaystyle = \frac{5}{14}$

$\displaystyle = 35,7$ %

The chances of drawing two red counters are:

$\displaystyle \frac{3}{8} \times \frac{2}{7}$

$\displaystyle = \frac{3}{28}$

$\displaystyle = 10,7$ %

So the chances of drawing either a green pair, or a red pair is:

$\displaystyle \frac{5}{14} + \frac{3}{28} $

$\displaystyle = \frac{13}{28}$

$\displaystyle = 46,4$ %