A) \[{{x}^{2}}+{{y}^{2}}-6x+4=0\]
B) \[{{x}^{2}}+{{y}^{2}}-3x+1=0\]
C) \[{{x}^{2}}+{{y}^{2}}-4y+2=0\]
D) \[{{x}^{2}}+{{y}^{2}}-6x-6y+10=0\]
Correct Answer: B
Solution :
Let A be the event of selecting a counterfeit coin and B be the event of getting head. Then \[P(A)=\frac{2}{16},\] \[P(\overline{A})=\frac{14}{16}\] and \[P\left( \frac{B}{A} \right)=1,\] \[P\left( \frac{B}{A} \right)=\frac{1}{2}\] Now, required probability \[=P(A\cap B)\cup P(\overline{A}\cap B)\] \[=P(A\cap B)+P(\overline{A}\cap B)\] \[=P(A)+P\left( \frac{B}{A} \right)+P(\overline{A})P\left( \frac{B}{A} \right)\] \[=\frac{2}{16}.1+\frac{14}{16}.\frac{1}{2}=\frac{9}{16}\]
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