A) \[\frac{3p+2{{p}^{2}}}{2}\]
B) \[\frac{p+3{{p}^{2}}}{4}\]
C) \[\frac{p+3{{p}^{2}}}{2}\]
D) \[\frac{3p+2{{p}^{2}}}{4}\]
Correct Answer: A
Solution :
| [a] We known that |
| P (exactly one of A or B occurs) |
| \[=P(A)+P(B)-2P(A\cap B)\] |
| \[\therefore P(A)+P(B)-2P(A\cap B)=p\] ? (1) |
| Similarly, \[P(B)+P(C)-2P(B\cap A)=p\] ? (2) |
| and \[P(C)+P(A)-2P(C\cap A)=p\] ? (3) |
| Adding Eqs. (1), (2), and (3) we get |
| \[2[P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A)]=3p\]\[\Rightarrow P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)\] |
| \[-P(C\cap A)]=3p/2\] ? (4) |
| It is also given that |
| \[P(A\cap B\cap C)={{p}^{2}}\] ? (5) |
| Now, |
| P(at least one of A, B and C) |
| \[=P(A)+p(B)+p(C)-p(A\cap B)-p(B\cap C)\] |
| \[-P(C\cap B)+P(A\cap B\cap C)\] |
| \[=\frac{3p}{2}+{{p}^{2}}=\frac{3p+2{{p}^{2}}}{2}\] [Using Eqs. (4) and (5)] |
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