probability of getting decoder rings in cereal boxes

tsh44

Junior Member
Joined
Sep 4, 2005
Messages
67
Hello, I having trouble with this problem.

One out of every five boxes of cereal has a secret message decoder ring. You buy 5 boxes hoping to get at least two rings. What are your chances.

The answer should be 821/3125.

Thanks for any help.
 
Re: probability

Hello, tsh44!

One out of every five boxes of cereal has a secret message decoder ring.
You buy 5 boxes hoping to get at least two rings. What are your chances?

The answer should be 821/3125.
You're expected to be familiar with Binomial Probability (or Independent Trials)
\(\displaystyle \;\;\)and its formula.
We are given: \(\displaystyle \,P(\text{ring})\,=\,\frac{1}{5},\;\;P(\text{no ring})\,=\,\frac{4}{5}\)

The opposite of "at least two" is "zero or one".

\(\displaystyle P(\text{0 rings})\;=\;C(5,0)\cdot\left(\frac{1}{5}\right)^o\left(\frac{4}{5}\right)^5\;=\;\frac{1024}{3125}\)

\(\displaystyle P(\text{1 ring})\;=\;C(5,1)\cdot\left(\frac{1}{5}\right)^1\left(\frac{4}{5}\right)^4\;=\;\frac{1280}{3125}\)

Hence: \(\displaystyle \,P(\text{0 or 1 ring})\;=\;\frac{1024}{3125}\,+\,\frac{1280}{3125}\;=\;\frac{2304}{3125}\)


Therefore: \(\displaystyle \,P(\text{2 or more rings})\;=\;1\,-\,\frac{2304}{3125}\;=\;\frac{821}{3125}\)
 
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