Re-sampling of a Histograms Bins
You may use Interpolation on the the bins and heights of the HistogramList to produce a smooth PDF for use in ProbabilityDistribution.  With a smooth distribution then you need not worry about increasing or decreasing bin sizes.
For example, with data
SeedRandom[9123]
data = RandomVariate[WeibullDistribution[2, 2], 10^6];

has Histogram
hist = Histogram[data, Automatic, &quot;PDF&quot;, PlotRange -&gt; Full]




You have the histogram bins and heights so I'll just use HistogramList to collect them and display with ListStepPlot.
hlData = HistogramList[data, Automatic, &quot;PDF&quot;];
lstep = ListStepPlot[Transpose@{First@hlData, Append[Last@hlData, 0]}, Mesh -&gt; Full]




A smooth InterpolatingFunction can be constructed from the bins and heights with Interpolation.
ifData = Interpolation[Transpose@{First@hlData, Append[Last@hlData, 0]}];
ifPlot = Plot[ifData[x], {x, Sequence @@ First@ifData[&quot;Domain&quot;]}, PlotStyle -&gt; Purple]




It can be seen that it matches the bins and heights.
Show[hist, ifPlot, lstep]




We want to use this function as the PDF of a ProbabilityDistribution so that NProbability can be calculated and RandomVariates can be generated.
The function needs to be non-negative over its domain.
NMinimize[ifData[x], {x} ∈ Interval @@ ifData[&quot;Domain&quot;]]


{0., {x -&gt; 7.4}}


Minimum value is zero at x equal to 7.4.
Must Integrate to 1 over its domain.
NIntegrate[ifData[x], {x} ∈ Interval @@ ifData[&quot;Domain&quot;]] // N


0.996821


The integral is just shy of 1 but we can ask ProbabilityDistribution to &quot;Normalize&quot; the PDF.
dist = ProbabilityDistribution[ifData[x], {x, Sequence @@ First@ifData[&quot;Domain&quot;]}, 
   Method -&gt; &quot;Normalize&quot;];

This distribution can be used in RandomVariate, Probability and other functions of the Random Variables guide.
Probabilities can be calculated
NProbability[0.9 &lt; x &lt; 2.2, x \[Distributed] dist]


0.514336


Pseudo-random numbers generated
RandomVariate[dist, 5]


{0.494373, 1.16545, 2.94366, 4.06116, 1.72519}


Properties like the SurvivalFunction, CDF, and others can be calculated.
Plot[SurvivalFunction[dist, x], {x, Sequence @@ First@ifData[&quot;Domain&quot;]}, 
 PlotStyle -&gt; Purple]




Update
Widening the bins will result in a loss of information so I would not recommend it.  However, if you must then Subdivide the InterpolatingFunction &quot;Domain&quot; into the number of bins you required.
downsamples =
  Function[numbins,
    With[{bins = Subdivide[##, numbins] &amp; @@ First@ifData[&quot;Domain&quot;]},
     {
      bins,
      NProbability[#1 &lt; x &lt; #2, x \[Distributed] dist] &amp; @@@ 
       Partition[bins, 2, 1]
      }]
    ] /@ Range[36, 12, -6];

Addressing your question in the comment; The total probability in each of the 5 cases is still one.
Total[downsamples[[All, 2]], {2}]


{1., 1., 1., 1., 1.}


However, notice the differences in the PDF histograms.
ListStepPlot[
 Transpose[{First@#, Append[Last@#, 0]}] &amp; /@ downsamples,
 Mesh -&gt; Full,
 PlotRange -&gt; All,
 PlotLegends -&gt; StringTemplate[&quot;`` bins&quot;] /@ Range[36, 12, -6]]




The initial histogram had 37 bins. The plot above shows how the information is lost as the bin width is widened (number of bins decreased).  I would recommend working with dist instead.
Hope this helps.