Probability: Harder Probability - Using Combinations
What is the probability of randomly picking a five-digit number whose leftmost digit is 1 and units digit is 9 from all the odd five-digit numbers that have distinct digits?
The numerator is the number of good choices (numbers with a leftmost digit of 1 and a units digit of 9 and that have 5 different digits). There is only one choice for the first and fifth digits (1 and 9, respectively), but the other digits can be any of the others, as long as there's no repetition. Therefore,
>$$\mathop{\fbox{${1}$}}\limits^{{1}{\text{st digit}}}\times
\mathop{\fbox{${8}$}}\limits^{{2}{\text{nd digit}}}\times
\mathop{\fbox{${7}$}}\limits^{{3}{\text{rd digit}}}\times
\mathop{\fbox{${6}$}}\limits^{{4}{\text{th digit}}}\times
\mathop{\fbox{${1}$}}\limits^{{5}{\text{th digit}}} $$.
Thus, your mighty fraction is
>$$\displaystyle \text{Probability} = \frac{1\times \require{cancel} \cancel{8} \times \require{cancel} \cancel{7} \times \require{cancel} \cancel{6} \times1}{8\times \require{cancel} \cancel{8} \times \require{cancel} \cancel{7} \times \require{cancel} \cancel{6} \times5}
=\frac{1}{40}$$.
Incorrect.
[[snippet]]
Incorrect.
[[snippet]]
Incorrect.
[[snippet]]
Incorrect.
[[snippet]]
Correct.
[[snippet]]
Start with the denominator—all odd five-digit numbers where the digits are all different. Break it down using __SeBoxes__:
>$$\mathop{\fbox{${}$}}\limits^{{1}{\text{st digit}}}\times
\mathop{\fbox{${}$}}\limits^{{2}{\text{nd digit}}}\times
\mathop{\fbox{${}$}}\limits^{{3}{\text{rd digit}}}\times
\mathop{\fbox{${}$}}\limits^{{4}{\text{th digit}}}\times
\mathop{\fbox{${}$}}\limits^{{5}{\text{th digit}}}$$.
Since the last digit has more constraints, start there:
* The fifth digit can be one of 5 odd digits (1, 3, 5, 7, and 9). Therefore, there are 5 choices for the fifth digit.
* The first digit can be one of 8 digits (there is no 0 in the beginning of a number, and the digit cannot be equal to the fifth digit). Therefore, there are 8 choices.
* The second digit can be one of 8 digits (the only constraint is that it cannot be the same as the first or fifth digits). Therefore, there are 8 choices.
* The third digit can be one of 7 digits (since 3 digits have already been used). Therefore, there are 7 choices.
* The fourth digit can be one of 6 digits (since 4 digits have already been used). Therefore, there are 6 choices.
The __SeBoxes__ are
>$$\mathop{\fbox{${8}$}}\limits^{{1}{\text{st digit}}}\times
\mathop{\fbox{${8}$}}\limits^{{2}{\text{nd digit}}}\times
\mathop{\fbox{${7}$}}\limits^{{3}{\text{rd digit}}}\times
\mathop{\fbox{${6}$}}\limits^{{4}{\text{th digit}}}\times
\mathop{\fbox{${5}$}}\limits^{{5}{\text{th digit}}}$$.
Calculating that big ol' number yet? Don't waste your time. First see to the numerator; you will probably be able to reduce that frightening number.
$$\frac{1}{80}$$
$$\frac{1}{40}$$
$$\frac{1}{25}$$
$$\frac{1}{20}$$
$$\frac{1}{10}$$
Continue
