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If A, B, and D are three events such that,\({\rm P}\left( {{\rm A} \cup {\rm B} \cup D} \right) = 0.7\)what is the value of\({\rm P}\left( {{{\rm A}^c} \cap {{\rm B}^c} \cap {D^c}} \right)\)?

Short Answer

Expert verified

The value of \({\rm P}\left( {{{\rm A}^c} \cap {{\rm B}^c} \cap {D^c}} \right) = 0.3\)

Step by step solution

01

Given information

Let A, B, D be the 3 events.

Thus,\({\rm P}\left( {{\rm A} \cup {\rm B} \cup D} \right) = 0.7\).

02

State the simple events and compute the probability

Formula for the probability of the intersection of complement of three arbitrary events is:\({\rm P}\left( {{{\rm A}^c} \cap {{\rm B}^c} \cap {D^c}} \right) = 1 - {\rm P}\left( {{\rm A} \cup {\rm B} \cup D} \right)\).

Therefore,

\(\begin{aligned}{}{\rm P}\left( {{{\rm A}^c} \cap {{\rm B}^c} \cap {D^c}} \right) &= 1 - {\rm P}\left( {{\rm A} \cup {\rm B} \cup D} \right)\\ &= 1 - 0.7\\ &= 0.3\end{aligned}\)

Therefore, the required probability is 0.3.

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