If I have nine cards, 3 red, 4 blue, and 2 purple, what is the probability of drawing two of the same card?

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Answer 1 · 2018-05-17T09:47:59

If your question is under that case "without putting the firct card back", the answer is $\frac{20}{72}$ $20/72$

There are three propabilities (getting two reds, getting two blues and getting two purples just after trial):

$P_{1} = \frac{3}{9} \times \frac{2}{8} = \frac{6}{72}$ $P_1 = 3/9 times 2/8 = 6/72$ (getting two reds)

$P_{2} = \frac{4}{9} \times \frac{3}{8} = \frac{12}{72}$ $P_2 = 4/9 times 3/8 = 12/72$ (getting two reds)

$P_{3} = \frac{2}{9} \times \frac{1}{8} = \frac{2}{72}$ $P_3 = 2/9 times 1/8 = 2/72$ (getting two purples).

Total probability is the sum of these

$P = \frac{6 + 12 + 2}{72} = \frac{20}{72}$ $P = (6+12+2)/72 = 20/72$

This is under the condition of when you pick the first card, you do not put it back to the cards.

Your answer is $\frac{20}{72} = \frac{10}{36} = \frac{5}{18}$ $20/72 = 10/36 = 5/18$