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Given the following probabilities, for what [tex]$P(B)$[/tex] are events [tex]$A$[/tex] and [tex][tex]$B$[/tex][/tex] independent?

[tex]
\begin{array}{c}
P(A)=\frac{1}{2} \\
P(A \text{ and } B)=\frac{1}{6}
\end{array}
[/tex]

A. [tex]\frac{3}{4}[/tex]
B. [tex]\frac{1}{3}[/tex]
C. [tex]\frac{1}{4}[/tex]
D. [tex]\frac{1}{8}[/tex]


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].
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