Question 1203532
.
Suppose that milk is shipped to retail outlets in boxes that hold 16 milk containers .
One particular box which happens to contain 6 underweight containers is opened for inspection 
and 5 containers are chosen at random. 
(a) {{{highlight(cross(Interpret))}}} <U>Find</U> the distribution of the number of underweight milk containers 
in the sample chosen by the inspector.
(b) Also {{{highlight(cross(comment))}}} <U>find</U> on the expected value value of the distribution.
(c) What is the probability that exact 2 underweight containers are selected
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The solution in the post by @Theo is &nbsp;INCORRECT. 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The &nbsp;CONCEPTUAL &nbsp;ERROR &nbsp;which he made is that this distribution &nbsp;IS &nbsp;BINOMIAL.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Actually, &nbsp;it is not binomial. &nbsp;&nbsp;See my correct solution below.



<pre>
(a) In part (a), the random variable X is the number of underweight containers in a sample of 5 containers;
    so, variable X may have 6 different values 0, 1, 2, 3, 4, 5.


    The probabilities are

    X= 0:  P(X=0) = {{{(10/16)*(9/15)*(8/14)*(7/13)*(6/12)}}} = 0.057692308   (5 not underweight containers)
                         choose 5 not underwait from 10 not underweight; 


    X= 1:  P(X=1) = {{{(10/16)*(9/15)*(8/14)*(7/13)*(1/6)}}}  = 0.096153846   (4 not underweight; 1 underweight)
                         choose 4 not underwait from 10 not underweight; choose 1 underweight from 6 underweight 


    X= 2:  P(X=2) = {{{(10/16)*(9/15)*(8/14)*(2/6)*(1/5)}}}   = 0.142857143   (3 not underweight; 2 underweight)
                         choose 3 not underwait from 10 not underweight; choose 2 underweight from 6 underweight 


    X= 3:  P(X=3) = {{{(10/16)*(9/15)*(3/6)*(2/5)*(1/4)}}}    = 0.187500000   (2 not underweight; 3 underweight)
                         choose 2 not underwait from 10 not underweight; choose 3 underweight from 6 underweight 


    X= 4:  P(X=4) = {{{(10/16)*(4/6)*(3/5)*(2/4)*(1/3)}}}     = 0.041666667   (1 not underweight; 4 underweight)
                         choose 1 not underwait from 10 not underweight; choose 4 underweight from 6 underweight 


    X= 5:  P(X=5) = {{{(5/6)*(4/5)*(3/4)*(2/3)*(1/2)}}}       = 0.277777778   (0 not underweight; 5 underweight)
                                                                         choose 5 underweight from 6 underweight 


(b) Expected value is

    E = 0*P(0) + 1 *P(1) + 2*P(2) + 3*P(3) + 4*(P4) + 5*P(5) = 

      = 0*0.057692308 + 1*0.096153846 + 2*0.142857143 + 3*0.018750000 + 4*0.041666667 + 5*277777778 = 2.499923687,

   Which after rounding is about E = 2.5.


(c)  See this value in (a) :  P(X=2) = 0.142857143,  or, after rounding,  0.1429.
</pre>

Solved.


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Do not use toooo smart words like &nbsp;" interpret " &nbsp;or &nbsp;" comment ".


It is &nbsp;NOT &nbsp;GOOD &nbsp;to use them in standard school problems.


Replace them by more typical words, &nbsp;as I did: &nbsp;these more typical words 
BETTER &nbsp;describe what the problem wants from a reader.