A) 0.06
B) 0.16
C) 0.84
D) 0.94
Correct Answer: C
Solution :
[c] Probability that machine stops working |
\[=P(A\cup B\cup C)\] |
\[\Rightarrow P(A\cup B\cup C)=P(A)+P(B)+P(C)\] |
\[-P(A\cap B)-P(A\cap C)-P(B\cap C)\]\[+P(A\cap B\cap C)\] |
\[\Rightarrow P(A\cup B\cup C)=P(A)+P(B)+P(C)\] |
\[-P(A)P(B)-P(A)P(C)\] |
\[-P(B)P(C)+P(A)P(B)P(C)\] |
(\[\because \] A, B & C are independent events) |
\[\Rightarrow P(A\cup B\cup C)=0.02+0.1+0.05-(0.02\times 0.1)\] |
\[-(0.02\times 0.05)-(0.1\times 0.05)\] |
\[+(0.02\times 0.05\times 0.1)\] |
\[\Rightarrow P(A\cup B\cup C)=0.16\] |
\[\therefore \] Probability that the machine will not stop working |
\[=1-P(A\cup B\cup C)=1-0.16=0.84\] |
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