Ratio

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neno89

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Oct 25, 2005
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Hello, can someone help me solve this problem? Thank you.

If two ratios are formed at random from the four numbers 1, 2, 4, and 8, what is the probability that the ratios are equal?
 
Does a given proportion have to use all four numbers, or can numbers be repeated?

Eliz.
 
4 cases when equal:
1/2 and 4/8
2/8 and 1/4
2/1 and 8/4
8/2 and 4/1

finish it? answer will be 4 / (all combos)

btw, what makes you think this is geometry?
 
I have seen this problem with the following interpretation.
But both Eliz and Dennis may be right?

If we can repeat a digit in numerator & denominator of the ratio, then can we think of these pairs
(1,1), (1,2), (1,4), (1,8)
(2,1), (2,2), (2,4), (2,8)
(4,1), (4,2), (4,4), (4,8)
(8,1), (8,2), (8,4), (8,8)
as (numerator, denominator) of the ratios?
Also, this is sampling with replacement. The two ratios are formed independently.
How many ratios equal to 1/8? Is it one?
What is the probability of two random ratios will both equal 1/8? Is it (1/16)(1/16)?

How many ratios equal to 2? Is it three?
What is the probability of two random ratios will both equal 2? Is it (3/16)(3/16)?

There are seven possible values for the ratios to equal. What are they?
Find the probability associated with each value.
What is the probability of two random ratios will both equal each one of those values? Then add them up.

If we cannot repeat remove the diagonal.
So P(1/8)=1/12 & P(2)=3/12 as opposed to 1/16 & 3/16.
Instead of seven possible values we now have six.
 
NENO89, can you give us any guidance on interpreting this problem?
Is there an answer given in the textbook? Please post it if given!
 
The answer in the back of the book is 1/3. but i don't understand how it was found.
 
Well then, Dennis is correct!
4/12 is 1/3.
 
ok I understand it now. Thank you soo much for those who helped!
 
pka said:
Well then, Dennis is correct!
4/12 is 1/3.
Click to expand...
Of course :roll:
(will I be interviewed and have to speak in 10 microphones at once?)
 
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