Let A and B be two independent events. The pr | vedclass

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(A) Given that $A$ and $B$ are independent events, $P(A \cap B) = P(A)P(B) = \frac{1}{12} \dots (1)$ The probability that neither $A$ nor $B$ happens

Vedclass · Accepted Answer

$P(A)=\frac{1}{3}, P(B)=\frac{1}{4}$

Vedclass · Answer

(A) Given that $A$ and $B$ are independent events, $P(A \cap B) = P(A)P(B) = \frac{1}{12} \dots (1)$ The probability that neither $A$ nor $B$ happens is $P(A' \cap B') = \frac{1}{2}$. By De Morgan's Law, $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) = \frac{1}{2}$, so $P(A \cup B) = \frac{1}{2}$. Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, we get: $P(A) + P(B) - \frac{1}{12} = \frac{1}{2} \implies P(A) + P(B) = \frac{1}{2} + \frac{1}{12} = \frac{7}{12} \dots (2)$ From $(1)$ and $(2)$, $P(A)$ and $P(B)$ are roots of the quadratic equation $x^2 - (P(A)+P(B))x + P(A)P(B) = 0$. $x^2 - \frac{7}{12}x + \frac{1}{12} = 0 \implies 12x^2 - 7x + 1 = 0$. Factoring the quadratic: $12x^2 - 4x - 3x + 1 = 0 \implies 4x(3x - 1) - 1(3x - 1) = 0 \implies (4x - 1)(3x - 1) = 0$. Thus, $x = \frac{1}{4}$ or $x = \frac{1}{3}$. Therefore, the probabilities are $P(A) = \frac{1}{3}$ and $P(B) = \frac{1}{4}$ (or vice versa).

Let $A$ and $B$ be two independent events. The probability that both $A$ and $B$ happen is $\frac{1}{12}$ and the probability that neither $A$ nor $B$ happens is $\frac{1}{2}$. Then

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