Probabilities

Tamara33

New member
Joined
Oct 30, 2010
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4
Hello everyone!

Here is the problem:
Michael is one of 7 racers that run on a 7 lane track. There will be two races. What is the probability that Michael will run in lane 1 at least ONCE in these two races?

At first I thought that I should add the probability of Michael randomly being assigned to lane 1 in both races (1/7 and 1/7), but the book says the answer is 13/49! I have no idea how to come up with this answer.

Any help would be GREATLY appreciated!

Thanks!
 
Try a binomial

\(\displaystyle \binom{2}{1}(\frac{1}{7})(\frac{6}{7})+\binom{2}{2}(\frac{1}{7})^{2}(\frac{6}{7})^{0}\)

OR

A good way to approach any "at least once" problem is to find the probability he does not run on lane 1 and subtract from 1.

\(\displaystyle 1-(\frac{6}{7})^{2}\)
 
Thank you very much!

Could you explain why the fraction is squared (in your seond example)? I'm trying to understand the theory behind this...
 
Because he did NOT get placed on the first lane in either race. The probability of NOT being placed in lane 1 is 6/7.

Since this happens twice, it is squared. Then, we subtract that from one because the problem asked for 'At least One".

Anytime you run into an "at least once" problem, think about the probability of it NOT happening and subtract from 1.

After all, the opposite of "at least one" is none.
 
Thanks, again, galactus! I'm totally with you when you discuss the probability of NOT getting into lane one. I was just confused about the reason for squaring this probability, rather than adding it to itself before subtracting it from one.

I really appreciate your help! This is a great site.
 
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