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Finite Math question....dont know how to do..><

tea217

Member
In a lottery game, you pick 6 numbers. Each # ranging from 1 to 49.
Then 6 #s are drawn...
What are the probability of three of ur hand picked numbers will match any 3 of the winning #s?? (Order doesn't matter..so it's a combination question)

Damn....i know how to get the answer to match 6 number...
Obviously it's 1/c(49,6) => 1/13983816
BUt..not with THREE matches. Please give me a full solution. Thx ^^

BTW...this game is call Lotto649....

 


<< Hrm, thats ages ago for me...
Isn't that like (49!6)/(49!3) ?
>>



Hmm, no it isn't 🙂

Ask me again when I wake up, didn't read the whole question at first.
 
Jackpot (all six winning numbers selected)
There are a total of 13,983,816 different groups of six numbers which could be drawn from the set {1, 2, ... , 49}. To see this we observe that there are 49 possibilities for the first number drawn, following which there are 48 possibilities for the second number, 47 for the third, 46 for the fourth, 45 for the fifth, and 44 for the sixth. If we multiply the numbers 49 x 48 x 47 x 46 x 45 x 44 we get 10,068,347,520. However, each possible group of six numbers (combination) can be drawn in different ways depending on which number in the group was drawn first, which was drawn second, and so on. There are 6 choices for the first, 5 for the second, 4 for the third, 3 for the fourth, 2 for the fifth, and 1 for the sixth. Multiply these numbers out to arrive at 6 x 5 x 4 x 3 x 2 x 1 = 720. We then need to divide 10,068,347,520 by 720 to arrive at the figure 13,983,816 as the number of different groups of six numbers (different picks). Since all numbers are assumed to be equally likely and since the probability of some number being drawn must be one, it follows that each pick of six numbers has a probability of 1/13,983,816 = 0.00000007151. This is roughly the same probability as obtaining 24 heads in succession when flipping a fair coin!

Second Prize (five winning numbers + bonus)
The pick of six must include 5 winning numbers plus the bonus. Since 5 of the six winning numbers must be picked, this means that one of the winning numbers must be excluded. There are six possibilities for the choice of excluded number and hence there are six ways for a pick of six to win the second place prize. The probability is thus 6/13,983,816 = 0.0000004291 which translates into odds against of 2,330,635:1.

Third Prize (five winning numbers selected, bonus number not selected)
As in the second prize there are six ways for a pick of six to include exactly five of the six drawn numbers. The remaining number must be one of the 42 numbers left over after the six winning numbers and the bonus number have been excluded. Thus there are a total of 6 x 42 = 252 ways for a pick of six to win the third prize. This becomes a probability of 252/13,983,816 = 0.00001802 or, equivalently, odds against of 55490.3:1.

Fourth Prize (four winning numbers selected)
There are 15 ways to include four of the six winning numbers and 903 ways to include two of the 43 non-winning numbers for a total of 15 x 903 = 13,545 ways for a pick of six to win the third prize, which works out to a probability of 13,545/13,983,816 = 0.0009686, that is odds against of 1031.4:1.

Fifth Prize (three winning numbers selected)
There are 20 ways to include three of the six winning numbers and 12,341 ways to include three of the 43 non-winning numbers for a total of 20 x 12,341 = 246,820 ways for a pick of six to win the fourth prize, which works out to a probability of 246,820/13,983,816 = 0.01765, that is odds against of 55.7:1.
 
Zeroburn..thx for your detailed answer. After reading your answer i remember how to do the question now. THanks again.
And you mention the ODDs against something... or Odds in favour of...

Well, my teacher taught us that before. But i am not too sure what does &quot;ODDS again/in favour of&quot; means. Can you tell me...detailedly.



<< Fifth Prize (three winning numbers selected)
There are 20 ways to include three of the six winning numbers and 12,341 ways to include three of the 43 non-winning numbers for a total of 20 x 12,341 = 246,820 ways for a pick of six to win the fourth prize, which works out to a probability of 246,820/13,983,816 = 0.01765, that is odds against of 55.7:1.
>>



My guess would be &quot;you will lose 55.7 times then you will win 1 time, then lose 55.7 times, win 1 time....&quot;
 
a is an event in a random experiment. the mathematical odds concerning a refer to the probability of a:

the odds in favor of a are ---> a : b
P(A) = a / (a + b)

the odds against a are ---> b : a
P(Ac) = b / (a + b).
 
I love Stats!

So how do you do it?

Do you find the probability of getting 1, 2, and 3 numbers and then adding the probabilities up?
 


<< the odds in favor of a are ---> a : b
P(A) = a / (a + b)

the odds against a are ---> b : a
P(Ac) = b / (a + b).
>>



I already know the crap about how to calculating ODDS......
but....i just dont see what this means.

damn....i suck. I can't grasp what &quot;odds against/in favour of&quot; mean....

 
Radiohead.
No..you don't find the probability of 1,2,3 #, then add it up.
You first find the total outcome...that is 13983816. (I think you know how to get that)

Then as for the event set A (only 3#s match):
c(6,3) = 20 ---> You've picked 6#s, now you've 3 of them to match only.
c(43,3) = 12341 ---> 43 non-winning #s, you need 3 more to make up 6#s (3winning+3nonwinning)
20 x 12341 = 246820 ---> Fundamental Counting Principle.....

P(A) = A / Totaloutcome
P(A) = 246820/13983816
 


<< I can't grasp what &quot;odds against/in favour of&quot; mean.... >>


Is this the type of thing you mean:

'Odds in favor' means how likely it is that a given circumstance will occur.
'Odds against' means how likely it is that a given circumstance will not occur.
 
Palad..


<< 'Odds in favor' means how likely it is that a given circumstance will occur. >>



This is what it confuses me when the teacher told me a definition close to this one.

&quot;how likely..&quot; This confuses me with probability.
I mean...&quot;the chance of occurrance&quot; is almost like &quot;how likely&quot;
Please clarify this for me. My english is not too good.

Also, say Odds in Favour of event A is 2:1. This means A is likely going to occur 2times and failing to occur once??




<< What math you in tea217?
-- mrcodedude
>>


I learned this in my gr13 finite math class. I guess another name for it is statistics &amp; probability.
Sigh..i am 18yr..and i am still suck at these things...
 
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