Question 1210330
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A grocer's bin contains 7 ripe tomatoes and 9 unripened tomatoes. 
Three tomatoes are selected at random (Tomatoes are not replaced). 
Show the probability distribution for the number of ripe tomatoes.
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<pre>
Let x be the number of ripe tomatoes in our random sample of 3 tomatoes.

They want you find and present the numbers  P(x=0),  P(x=1),  P(x=2)  and  P(x=3).



(0)  x=0  means that in our sample all 3 tomatoes are unripped; there is no ripped tomatoes.

     So, P(x=0) = {{{C[9]^3/C[16]^3}}} = {{{84/560}}} = {{{3/20}}}.

     Here  {{{C[16]^3}}} = 560 in the denominator is the total number of triples of 16 = 7+9 tomatoes.

           {{{C[9]^3}}} = 84 in the numerator is the number of triples of 9 unripped tomatoes.




(1)  x = 1  means that in our sample 1 tomato is ripped; two other tomatoes are unripped.

     So, P(x=1) = {{{(C[9]^2*C[7]^1)/C[16]^3}}} = {{{(36*7)/560}}} = {{{252/560}}} = {{{9/20}}}.

     Again, here  {{{C[16]^3}}} = 560 in the denominator is the total number of triples of 16 = 7+9 tomatoes.

           {{{C[9]^2*C[7]^1}}} = 36*7 = 252 in the numerator is the number of triples, 
                                      consisting of 2 unripped tomatoes and 1 ripped tomato.




(2)  x = 2  means that in our sample 2 tomatoes are ripped; one tomato is unripped.

     So, P(x=2) = {{{(C[9]^1*C[7]^2)/C[16]^3}}} = {{{(9*21)/560}}} = {{{189/560}}} = {{{27/80}}}.

     Again, here  {{{C[16]^3}}} = 560 in the denominator is the total number of triples of 16 = 7+9 tomatoes.

           {{{C[9]^1*C[7]^2}}} = 9*21 = 189 in the numerator is the number of triples, 
                                      consisting of 1 unripped tomato and 2 ripped tomatoes.



(3)  x = 3  means that in our sample 3 tomatoes are ripped; no unripped tomatoes.

     So, P(x=3) = {{{(C[9]^0*C[7]^3)/C[16]^3}}} = {{{(1*35)/560}}} = {{{35/560}}} = {{{1/16}}}   after reducing.

     Again, here  {{{C[16]^3}}} = 560 in the denominator is the total number of triples of 16 = 7+9 tomatoes.

           {{{C[9]^0*C[7]^3}}} = 1*35 = 35 in the denominator is the number of triples, 
                                      consisting of 0 unripped tomatoes and 3 ripped tomatoes.


Thus  P(0) = {{{3/20}}};  P(1) = {{{9/20}}};  P(2) = {{{27/80}}};  P(3) = {{{1/16}}} = {{{5/80}}}.


The sum  P(0) + P(1) + P(2) + P(3) = {{{3/20}}} + {{{9/20}}} + {{{27/80}}} + {{{5/80}}} = {{{12/20}}} + {{{32/80}}} = {{{48/80}}} + {{{32/80}}} = {{{80/80}}} = 1.
</pre>

Solved.