Question 1194951
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Neither problem is clearly defined.  I will assume....<br>
(1) without replacement
(2) order does not matter<br>
(a) a club and a diamond<br>
On the first draw, there are 52 cards to choose from, of which 26 are okay (13 clubs and 13 diamonds).  P = 26/52 = 1/2.<br>
On the second draw, there are 51 cards remaining, of which 13 are okay (the 13 of the suit that was not drawn on the first draw).  P = 13/51.<br>
ANSWER: (1/2)(13/51) = 13/102<br>
(b) one card of each suit<br>
1st draw: 52 cards; can choose any.  P = 52/52 = 1
2nd draw: 51 cards remaining; can choose any of the 39 in the 3 remaining suits.  P = 39/51 = 13/17
3rd draw: 50 cards remaining; can choose any of the 26 cards in the 2 remaining suits.  P = 26/50 = 13/25
4th draw: 49 cards remaining; must choose one of the 13 cards in the last remaining suit.  P = 13/49<br>
ANSWER: (1)(13/17)(13/25)(13/49)<br>
Use a calculator and express the result in the desired/required form.<br>