Q67 / 180 · KCET 2021 · Mathematics
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$3 P-2 P^{2}$
- Correct answerC$3 P^{2}-2 P^{3}$
- D$3 P^{2}$
Solution
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 $3 P^{2}-2 P^{3}$.
Hence, the correct answer is option: 'C'.