# If the chance of an event was 1 <mrow class="MJX-TeXAtom-ORD"> / </mrow> 128 and

If the chance of an event was $1/128$ and increased by 20%, what is the new chance?
So I have something that has a 1/128 chance of occurring, let's say. Suddenly, the chances of that thing happening are increased by 20%. How is that fraction written? Would you multiply 1/128 by 6/5 (yielding 3/320), or would you take 128, multiply it by 4/5, and then invert it (1/102)?
This isn't for homework, I'm merely curious.
Thank you
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Mateo Carson
Multiplying by $6/5=1.2$ is the correct approach. This increases a value by $20\mathrm{%}$. Multiplying by $4/5$ decreases a value by $20\mathrm{%}$, but inverting that is not the same as increasing by $20\mathrm{%}$. It results in multiplying by $5/4$, which is an increase of $25\mathrm{%}$. On top of that, $128\cdot \frac{4}{5}=102.4,$ not $102$

ddcon4r
The first approach is correct.
Initial $p=\frac{1}{128}$, you need to increase $p$ by $20\mathrm{%}$
Note
Since $p=\frac{1}{128}$, i.e. in the form of a fraction, it is possible to obtain the correct result by manipulating its denominator only, but then you would need to divide the denominator by $1.2$i.e. multiply it by $\frac{5}{6}$