Question 1183431
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Three yam tubers are chosen at random from 15 tubers of which 5 are spoilt. Find the probability that, of the three chosen tubers:
A. None is spoilt
B. All are spoilt
C. Exactly one is spoilt
D. At least one is spoilt.
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(A)  In part (A), you chose, actually, 3 tubers from 10 tubers that are not spoilt, among the total of 15 tubers.


     Therefore, you can use any of two formulas to find the probability


         P = {{{(10/15)*(9/14)*(8/13)}}} = {{{(2/3)*(9/14)*(8/13)}}} = {{{24/91}}} = 0.2637  (rounded),

     or

         P = {{{C[10]^3/C[15]^3}}} = {{{120/455}}} = {{{24/91}}} = 0.2637  (same as above).




(B)  In part (B), you select, actually, 3 tubers from 5 spoilt tubers, among the total 15 tubers.


     Therefore, you can use any of two formulas to find the probability


         P = {{{(5/15)*(4/14)*(3/13)}}} = {{{(1/3)*(2/7)*(3/13)}}} = {{{2/91}}} = 0.0219  (rounded),

     or

         P = {{{C[5]^3/C[15]^3}}} = {{{10/455}}} = {{{2/91}}} = 0.0219  (the same value).



(C)  "Exactly one is spoilt" means one is spoilt of the 3 selected among the total 15, and 2 are not.


     Therefore, you can use this formula to find the probability


         P = {{{(C[5]^1*C[10]^2)/C[15]^3}}} = {{{(5*45)/455}}} = {{{45/91}}} = 0.4945.    (rounded)



(D)  It is the complement probability to the value found in (A)


         P = 1 - 0.2637 = 0.7363   (rounded).
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