probability--registered runners in a marathon

kristinel

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Aug 22, 2007
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Here's a question I'm pondering today.

A running club found that 20% of runners withdraw from a marathon without completing the race. 20 runners have registered for a mini-marathon.

What is the probability that more than three runners will withdraw?

I think I need to use a binomial distribution function. My question is, how do I treat the 'more than three' part of the question? If 20% of 20 runners drop out, that means we think that four runners will drop out, correct? So, do I use the binomial function for the values (3) and (4), and add them together? I just wanted to make sure I account for all parts of the question. Any ideas?
 
You're correct. For "more than 3", add them up from 4 to 20.

\(\displaystyle \L\\\sum_{k=4}^{20}C(20,k)(0.20)^{k}(0.80)^{20-k}\)
 
Hello, kristinel!

A running club found that 20% of runners withdraw from a marathon
without completing the race. 20 runners have registered for a mini-marathon.

What is the probability that more than three runners will withdraw?

You're right . . . it's a binomial distribution.

As Galactus pointed out:
. . "More than 3" means: \(\displaystyle \:\text{4 or 5 or 6 or 7 or . . . 19 or 20}\)
. . You must find the separate probabilities and add them.


Or we can find the opposite: \(\displaystyle \:\text{3 or less\) . . . and subtract from \(\displaystyle 1\).

\(\displaystyle \begin{array}{ccccc}P(0) & = & {20\choose0}(0.2)^0(0.8)^{20} & = & 0.011529215 \\
P(1) & = & {20\choose1}(0.2)^1(0.8)^{19} & = & 0.057646075 \\
P(2) & = & {20\choose2}(0.2)^2(0.8)^{18} & = & 0.136909429 \\
P(3) & = & {20\choose3}(0.2)^3(0.8)^{17} & = & 0.205364143 \\ \hline
& & \text{Total:} & & 0.411448862
\end{array}\)


Therefore: \(\displaystyle \:p(\text{more than 3}) \;=\;1\,-\,0.411448962\;=\;0.588551138\)

 
Thank you both so much for your explanations. That makes much more sense to me now :D
 
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