| Direction: In each of the following questions read the given statements and compare the given two quantities on its basis. |
| A box contains 6 pink balls, 5 blue balls, 3 green balls and 6 yellow balls. |
| Quantity I. If three balls are drawn at random then the probability that all balls are either blue or yellow. |
| Quantity II. If three balls are drawn at random then the probability that all are of different colours. |
A) Quantity I > Quantity II
B) Quantity I \[\ge \] Quantity II
C) Quantity I \[\le \] Quantity II
D) Quantity I < Quantity II
E) No relation between Quantity I and II
Correct Answer: D
Solution :
Total no. of balls \[=6+5+3+6=20\] Quantity I. \[n(S)={}^{20}{{C}_{3}}\] \[\therefore \,\,n(E)={}^{5}{{C}_{3}}+{}^{6}{{C}_{3}}\] \[\therefore \,\,P(E)=\frac{{}^{5}{{C}_{3}}+{}^{6}{{C}_{3}}}{{}^{20}{{C}_{3}}}=\frac{30}{1140}=\frac{3}{114}\] Quantity II. \[n(S)={}^{20}{{C}_{3}}=1140\] \[n(E)={}^{6}{{C}_{1}}\times {}^{5}{{C}_{1}}\times {}^{3}{{C}_{1}}+{}^{5}{{C}_{1}}\times {}^{3}{{C}_{1}}\times {}^{6}{{C}_{1}}\] \[+{}^{3}{{C}_{1}}\times {}^{6}{{C}_{1}}\times {}^{6}{{C}_{1}}+{}^{6}{{C}_{1}}\times {}^{5}{{C}_{1}}\times {}^{6}{{C}_{1}}\] \[n(E)=6\times 5\times 3+5\times 3\times 6+3\times 6\times 6+6\times 5\times 6\] \[=90+90+108+180=468\] \[\therefore \] \[P(E)=\frac{468}{1140}\] Hence Quantity I < Quantity II
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