KCET · Maths · Probability
If \(A, B\) and \(C\) are three independent events such that \(P(A)=P(B)=P(C)=P\), then \(P\) (at least two of \(A, B\) and \(C\) occur) is equal to
- A \(P^{3}-3 P\)
- B \(3 P-2 P^{2}\)
- C \(3 P^{2}-2 P^{3}\)
- D \(3 P^{2}\)
Answer & Solution
Correct Answer
(C) \(3 P^{2}-2 P^{3}\)
Step-by-step Solution
Detailed explanation
Given, \(P(A)=P(B)=P(C)=P\)
\(P\) (At least two of \(A, B, C\) occur)
\(=\left(P \cap B \cap C^{\prime}\right)+P\left(A \cap B^{\prime} \cap C\right)+P\left(A^{\prime} \cap B \cap C\right)\)
\(+P(A \cap B \cap C)\)
\(=P(A) P(B) P\left(C^{\prime}\right)+P(A) P\left(B^{\prime}\right) P(C)+P\left(A^{\prime}\right) P(B) P(C)\)
\(+P(A) P(B) P(C)\)
\(=P(A) P(B)[1-P(C)]+P(A)[1-(B)] P(C)\)
\(+[1-P(A)] P(B) P(C)+P(A) P(B) P(C)\)
\(=P \times P \times(1-P)+(1-P) \times P \times P+P \times(1-P)\)
\(\times P+P \times P \times P\)
\(=P^{2}[(1-P)+(1-P)+(1-P)+P]\)
\(=P^{2}[3-2 P]\)
\(=3 P^{2}-2 P^{3}\)
\(P\left(\right.\) At least two of \(A, B, C\) occur) is \(3 P^{2}-2 P^{3}\).
\(P\) (At least two of \(A, B, C\) occur)
\(=\left(P \cap B \cap C^{\prime}\right)+P\left(A \cap B^{\prime} \cap C\right)+P\left(A^{\prime} \cap B \cap C\right)\)
\(+P(A \cap B \cap C)\)
\(=P(A) P(B) P\left(C^{\prime}\right)+P(A) P\left(B^{\prime}\right) P(C)+P\left(A^{\prime}\right) P(B) P(C)\)
\(+P(A) P(B) P(C)\)
\(=P(A) P(B)[1-P(C)]+P(A)[1-(B)] P(C)\)
\(+[1-P(A)] P(B) P(C)+P(A) P(B) P(C)\)
\(=P \times P \times(1-P)+(1-P) \times P \times P+P \times(1-P)\)
\(\times P+P \times P \times P\)
\(=P^{2}[(1-P)+(1-P)+(1-P)+P]\)
\(=P^{2}[3-2 P]\)
\(=3 P^{2}-2 P^{3}\)
\(P\left(\right.\) At least two of \(A, B, C\) occur) is \(3 P^{2}-2 P^{3}\).
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