Question 149493
In a shop there are 20 customers, 18 of whom will make a purchase. If three customers are selected, one at a time, at random, what is the probability that all will make a purchase? Is it 0.7717, 0.7605, 0.8524, or 0.8808. 
Please help 
<pre><font size = 3 color = "indigo"><b>

It's not any of those.  It's 0.7158

P(1st chosen will purchase 

      <font size = 5>AND</font> 

           2nd chosen will purchase 

                  <font size = 8>AND</font>

                      3rd chosen will purchase)

                            <font size = 8>=</font>

P(1st chosen will purchase) 

<font size = 8>TIMES</font>

P(2nd chosen will purchase, given that the 1st chosen will purchase)

<font size = 8>TIMES</font>

P(3rd chosen will purchase, given that the 1st and 2nd chosen will purchase)

<font size = 8>=</font>

{{{ (18/20)(17/19)(16/18) = 68/95 = 0.7157894737}}} or rounded to ten-thousandths,

{{{0.7158}}}.

You can also do it using combinations: 

{{{Number_of_choices_of_3_purchasers_out_of_the_18_purchasers/Number_of_choices_of_3_customers_out_of_the_20_customers}}}

<font size = 5>=

{{{(C(18,3))/(C(20,3))  = 816/1140 = 68/95 = 0.7157894737}}} 

Stanbon's solution is incorrect because 
the choices are not independent. Binomial
probabilities are ONLY for independent choices.
The choices are not independent because the 
probability of selecting the second one as a 
purchaser CHANGES after selecting the first one. 
Also the probability of selecting the third one 
as a purchaser CHANGES after selecting the 
first two.

</font>Edwin</pre></b>