Question

You want to make five-letter codes that use the letters A, F, E, R, and M without repeating any letter. What is the probability that a randomly chosen code starts with M and ends with E?


0.01



0.02



0.05



0.10

Answer 1

\(\Huge{\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}}\\\color{white}{.}\\\Huge\sf\color{blue}{~~~~Welcome~to~OpenStudy!~\ddot\smile}\\\color{white}{.}\\\\\Huge{\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}}\)

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Answer 2

This is a pretty simple problem ^^

Now we have \(\large{5}\) letters that we would want to make into a word that does not repeat but ends with an \(\large{m}\) and an \(\large{n}\) so we determine how many letters there are that are m out of the \(\large{5}\) letters....

Well we have \(\Large{\frac{1}{5}}\) of the letters is an \(\large{m}\) now we determine the fraction of how many letters is an \(\large{n}\)....

Since we know that one is an \(\large{m}\) we would take away a letter and compare with only \(\large{4}\) letters...So there is \(\Large{\frac{1}{4}}\) of the letters that is an \(\large{n}\)....

Now that we know this we would multiply....

\(\huge\color{purple}{\frac{1}{5}\times \frac{1}{4}=probability}\)

Answer 3

sorry it took so long...

Answer 4

@banug

Answer 5

thank you.

Answer 6

\(\Huge\cal\rlap{\color{red}{No~Problem!:)}}{\color{blue}{\;No~Problem!:)}}\)