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Nine cards 1-9 are in a box and two cards are selected without replacement find probablity that both numbers selected are odd even though their sum is even?
Ametzcher
10
Aracely Wolff
Wiki User
∙ 14y agoSince the selected cards are not replaced, the selections are dependent.
It is not given how many cards have an odd or even number. However, there are 2, 4, or 6 cards with an odd number, since the sum of all 9 card numbers is even.
Let say that would be 4 cards with odd numbers, and 5 cards with even numbers.
Then the probability that the first card selected with an odd number is 4/9.
Now there are 8 cards left, of which 3 are with odd numbers, so the probability that the second card selected is a card with an odd number is 3/8.
Therefore, the probability that the first and the second cards selected are with odd numbers is 4/9 x 3/8 = 12/72 = 3/18
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