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Sagot :
To determine [tex]\( P(B) \)[/tex] such that events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are independent given the probabilities:
[tex]\[ \begin{array}{c} P(A) = \frac{1}{2} \\ P(A \text{ and } B) = \frac{1}{6} \end{array} \][/tex]
we need to use the definition of independence for two events. Events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are independent if and only if:
[tex]\[ P(A \text{ and } B) = P(A) \cdot P(B) \][/tex]
Given:
[tex]\[ P(A) = \frac{1}{2} \][/tex]
[tex]\[ P(A \text{ and } B) = \frac{1}{6} \][/tex]
We need to find [tex]\( P(B) \)[/tex] such that the equation holds true. Substitute the given values into the equation:
[tex]\[ \frac{1}{6} = \frac{1}{2} \cdot P(B) \][/tex]
To solve for [tex]\( P(B) \)[/tex], we can multiply both sides of the equation by 2:
[tex]\[ P(B) = \frac{1}{6} \cdot 2 \][/tex]
Perform the multiplication:
[tex]\[ P(B) = \frac{2}{6} = \frac{1}{3} \][/tex]
Therefore, the value of [tex]\( P(B) \)[/tex] that makes events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] independent is:
[tex]\[ \boxed{\frac{1}{3}} \][/tex]
So, [tex]\( P(B) = \frac{1}{3} \)[/tex].
[tex]\[ \begin{array}{c} P(A) = \frac{1}{2} \\ P(A \text{ and } B) = \frac{1}{6} \end{array} \][/tex]
we need to use the definition of independence for two events. Events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are independent if and only if:
[tex]\[ P(A \text{ and } B) = P(A) \cdot P(B) \][/tex]
Given:
[tex]\[ P(A) = \frac{1}{2} \][/tex]
[tex]\[ P(A \text{ and } B) = \frac{1}{6} \][/tex]
We need to find [tex]\( P(B) \)[/tex] such that the equation holds true. Substitute the given values into the equation:
[tex]\[ \frac{1}{6} = \frac{1}{2} \cdot P(B) \][/tex]
To solve for [tex]\( P(B) \)[/tex], we can multiply both sides of the equation by 2:
[tex]\[ P(B) = \frac{1}{6} \cdot 2 \][/tex]
Perform the multiplication:
[tex]\[ P(B) = \frac{2}{6} = \frac{1}{3} \][/tex]
Therefore, the value of [tex]\( P(B) \)[/tex] that makes events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] independent is:
[tex]\[ \boxed{\frac{1}{3}} \][/tex]
So, [tex]\( P(B) = \frac{1}{3} \)[/tex].
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