Question 1149232
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X=1: You need to get a counterfeit on the first draw.
{{{P(1) = 4/10}}}<br>
X=2: You need to get a genuine first and a counterfeit second.
{{{P(2) = (6/10)(4/9)}}}<br>
X=3: You need to get a genuine on the first two draws and a counterfeit on the third.
{{{P(3) = (6/10)(5/9)(4/8)}}}<br>
Similarly....<br>
{{{P(4) = (6/10)(5/9)(4/8)(4/7)}}}
{{{P(5) = (6/10)(5/9)(4/8)(3/7)(4/6)}}}
{{{P(6) = (6/10)(5/9)(4/8)(3/7)(2/6)(4/5)}}}
{{{P(7) = (6/10)(5/9)(4/8)(3/7)(2/6)(1/5)(4/4)}}}<br>
Simplify as required to find the probability density function.<br>
It is interesting to observe that the sum of all the probabilities is indeed 1.<br>
The sample space is 1 to 7.  Clearly if there are 6 genuine coins and 4 counterfeit coins, the "worst case" if you are trying to get a counterfeit coin is to draw all 6 of the genuine coins first, thus drawing the first counterfeit coin on the 7th draw.<br>