A) \[\frac{11}{40}\]
B) \[\frac{3}{10}\]
C) \[\frac{1}{40}\]
D) none of these.
Correct Answer: A
Solution :
[a] : Let us define the events in the following way: A : 4 being the minimum number B : 8 being the maximum number \[A\cap B:1\] being the minimum number and 8 being the maximum number Therefore \[P(A)=\frac{^{6}{{C}_{2}}}{^{10}{{C}_{3}}}=\frac{15}{120}\] \[P(B)=\frac{^{7}{{C}_{2}}}{^{10}{{C}_{3}}}=\frac{21}{120}\]and \[P(A\cap B)=\frac{^{3}{{C}_{2}}}{^{10}{{C}_{3}}}=\frac{3}{120}\] \[\therefore \]The required probability, \[P(A\cup B)=P(A)+P(B)-P(A\cap B)\] \[=\frac{15}{120}+\frac{21}{120}-\frac{3}{120}=\frac{33}{120}=\frac{11}{40}\]
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