Question 425873: Hello,
I have tried to solve a probability problem and have not reach any good answer. I would like someone to help me with this problem. Thank you!
Problem:
The numbers 1 through 9 are put into a hat. What is the probability that all odd numbers will be chosen first if one card at a time is drawn from the hat and not replaced?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you have the number 1 through 9.
5 of those numbers are odd (1,3,5,7,9)
4 of those numbers are even (2,4,6,8)
if he wants to draw all odd cards from the deck, then it will require a minimum of 5 draws from the deck.
the probability that the first card is odd is equal to 5/9
the probability that the second card is odd is equal to 4/8
the probability that the third card is odd is equal to 3/7
the probability that the fourth card is odd is equal to 2/6
the probability that the fifth card is odd is equal to 1/5
the probability that all 5 odd cards will be drawn from the deck in 5 draws from the deck is:
5/9 * 4/8 * 3/7 * 2/6 * 1/5
the total probability becomes .007936508
to see how this works, use smaller numbers.
suppose you have 3 numbers (1,2,3)
p(getting all odd) would require a minimum of 2 draws.
p(odd on first draw) = 2/3
p(odd on second draw) = 1/2
p(odd on first and second draw) = 2/6
here are the possible combination when you draw 2 cards out of 3.
first draw second draw combination
1 2 1 odd 1 even
2 1 1 even 1 odd
1 3 2 odd *****
3 1 2 odd *****
2 3 1 even 1 odd
3 2 1 odd 1 even
only 2 out of 6 possible combinations have all odd numbers.
same idea works with your problem.
answer for your problem is:
(5*4*3*2*1) / (9*8*7*6*5) = .007936508
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