Question 479505
Nine tickets, numbered 1 to 9, are in a box. If 2 tickets are drawn, determine the probability p that one is odd and one is even. 
Here is what I did (Method 1): 
P(1=even, 9 tickets)
n (total) = 9, n (even) = 5, n (odd) = 4
P(1=odd, 9 tickets) 
P (A and B) = P(A) times P(B)
[(1/9)X4]X[(1/8)X5] = 5/18 However, my textbook's answer is 5/9. 
Even if I draw even ticket first, then odd ticket, I will get 5/9 since:
[(1/9)X5]X[(1/8)times4] = 5/18. 
Method 2:
5C1/9C1 = 5/9
4C1/8C1 = 1/2
P(A and B) = 5/9 X 1/2 = 5/18. While it confirms my answer obtained through Method 1, the textbook's answer is 5/9. 
Prior solving this problem, I thought about it this way: 
We have 9 total tickets, 5 are even, and 4 are odd. 
Since we are dealing with tickets, common sense tells me 1. we are taking one ticket at a time; 2. tickets cannot be replaced; 3. after I draw my first ticket, I will have one less ticket and will draw my second ticket from smaller number of tickets, thus we have dependant events; 4. combinations is one way to solve this problem; 5. multiply probability of Event 1 with probability of Event 2 to get total probability. 
Here is how the textbook did it: 
(5C1 X 4C1) / 9C2 = 5/9. 
Please note how they are drawing first ticket and second ticket from 9 tickets as if cards WERE REPLACED. WHY? 
Please let me know where I went wrong in my reasoning. 
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Nine tickets, numbered 1 to 9, are in a box. If 2 tickets are drawn, determine the probability p that one is odd and one is even. 
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There are 3 possible results: 2 odd ; 2 even ; 1 odd and 1 even
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# of even digits = 4
# of odd digits = 5
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P(2 even) = 4C2/9C2 = 6/36 = 1/6
P(2 odd) = 5C2/9C2 = 10/36 = 5/18
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P(one odd and one even) 
= 1 - [P(2 even)+P(2 odd)] 
= 1-[(3/18)+5(18)] 
= 10/18
= 5/9
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Cheers,
Stan H.
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