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The sum of probabilities in a probability distribution must equal $1$.$$\sum P(X) = 1$$$$\frac{k}{2} + \frac{k}{3} + \frac{k}{6} = 1$$Find the common denominator (which is $6$):$$\frac{3k}{6} + \frac{2k}{6} + \frac{1k}{6} = 1$$$$\frac{6k}{6} = 1$$$$k = 1$$(ii) Find $P(1 \le X < 3)$The range $1 \le X < 3$ includes $X=1$ and $X=2$, but not $X=3$.$$P(1 \le X < 3) = P(X=1) + P(X=2)$$Substituting $k=1$:$$P(X=1) = \frac{1}{2}, \quad P(X=2) = \frac{1}{3}$$$$P(1 \le X < 3) = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$$(iii) Find $E(X)$, the mean of $X$The expectation (mean) is defined as $E(X) = \sum X \cdot P(X)$.Using $k=1$:$$P(X=1) = \frac{1}{2}, \quad P(X=2) = \frac{1}{3}, \quad P(X=3) = \frac{1}{6}$$$$E(X) = \left(1 \times \frac{1}{2}\right) + \left(2 \times \frac{1}{3}\right) + \left(3 \times \frac{1}{6}\right)$$$$E(X) = \frac{1}{2} + \frac{2}{3} + \frac{3}{6}$$$$E(X) = \frac{1}{2} + \frac{2}{3} + \frac{1}{2}$$$$E(X) = 1 + \frac{2}{3} = \frac{5}{3}$$OR Part: Independent EventsProblem:$A$ and $B$ are independent events such that:$P(A \cap \overline{B}) = \frac{1}{4}$$P(\overline{A} \cap B) = \frac{1}{6}$Solution:Let $P(A) = x$ and $P(B) = y$.Since events are independent:$$P(A \cap \overline{B}) = P(A) \cdot P(\overline{B}) = x(1-y)$$$$P(\overline{A} \cap B) = P(\overline{A}) \cdot P(B) = (1-x)y$$We have the system of equations:$x(1-y) = \frac{1}{4} \implies x - xy = \frac{1}{4}$$(1-x)y = \frac{1}{6} \implies y - xy = \frac{1}{6}$Subtracting equation (2) from equation (1):$$(x - xy) - (y - xy) = \frac{1}{4} - \frac{1}{6}$$$$x - y = \frac{3}{12} - \frac{2}{12} = \frac{1}{12}$$$$x = y + \frac{1}{12}$$Substitute $x$ back into equation (2):$$\left(1 - \left(y + \frac{1}{12}\right)\right)y = \frac{1}{6}$$$$\left(\frac{11}{12} - y\right)y = \frac{1}{6}$$$$\frac{11y}{12} - y^2 = \frac{1}{6}$$Multiply by 12 to clear the denominator:$$11y - 12y^2 = 2$$$$12y^2 - 11y + 2 = 0$$Solving the quadratic equation for $y$:$$y = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(12)(2)}}{2(12)}$$$$y = \frac{11 \pm \sqrt{121 - 96}}{24} = \frac{11 \pm \sqrt{25}}{24} = \frac{11 \pm 5}{24}$$Two possible values for $y$:$y_1 = \frac{16}{24} = \frac{2}{3}$$y_2 = \frac{6}{24} = \frac{1}{4}$Finding corresponding $x$ values using $x = y + \frac{1}{12}$:If $y = \frac{2}{3}$: $x = \frac{2}{3} + \frac{1}{12} = \frac{8+1}{12} = \frac{9}{12} = \frac{3}{4}$If $y = \frac{1}{4}$: $x = \frac{1}{4} + \frac{1}{12} = \frac{3+1}{12} = \frac{4}{12} = \frac{1}{3}$Answer:There are two possible sets of values:$P(A) = \frac{3}{4}$ and $P(B) = \frac{2}{3}$$P(A) = \frac{1}{3}$ and $P(B) = \frac{1}{4}$

If events $E$ and $F$ are independent, then $$\mathbf{P(E \mid F) = P(E)}$$And similarly, if $E$ and $F$ are independent, then $E$ and $\overline{F}$ are also independent:$$\mathbf{P(E \mid \overline{F}) = P(E)}$$ Substitute the Given ValueSince $E$ and $F$ are independent events, we can directly state the result:$$P(E \mid \overline{F}) = P(E)$$Given that $P(E) = \frac{2}{3}$, we have:$$P(E \mid \overline{F}) = \frac{2}{3}$$

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