We are given two independent events $A$ and $B$ such that

$\Pr(A \cap B) = \tfrac{1}{8}$.

$\Pr(A^c \cap B^c) = \tfrac{3}{8}$.

Let $\Pr(A) = p$ and $\Pr(B) = q$. Since $A$ and $B$ are independent,

$ \Pr(A \cap B) \;=\; p \times q \;=\; \tfrac{1}{8}. \tag{1} $

Also, by independence of complements,

$ \Pr(A^c \cap B^c) \;=\; \Pr(A^c)\,\Pr(B^c) \;=\; (1 - p)(1 - q) \;=\; \tfrac{3}{8}. \tag{2} $

Step 1: Express $\Pr(A^c \cap B^c)$ in terms of $p$ and $q$

Expanding (2):

$ (1 - p)(1 - q) \;=\; 1 - p - q + p\,q \;=\; \tfrac{3}{8}. $

Since from (1) we have $p\,q = \tfrac{1}{8}$, substitute that:

$ 1 - p - q + \tfrac{1}{8} = \tfrac{3}{8} \;\;\Longrightarrow\;\; 1 - p - q = \tfrac{3}{8} - \tfrac{1}{8} = \tfrac{2}{8} = \tfrac{1}{4}. $

Hence,

$ -\,p \;-\; q = \tfrac{1}{4} - 1 = -\tfrac{3}{4} \quad\Longrightarrow\quad p + q = \tfrac{3}{4}. \tag{3} $

Step 2: Solve the system

We have:

$ \begin{cases} p + q = \tfrac{3}{4},\\[6pt] p\,q = \tfrac{1}{8}. \end{cases} $

From $p + q = \tfrac{3}{4}$, let $q = \tfrac{3}{4} - p$. Substituting into $p\,q = \tfrac{1}{8}$,

$ p \,\bigl(\tfrac{3}{4} - p\bigr) = \tfrac{1}{8} \;\;\Longrightarrow\;\; \tfrac{3}{4}\,p - p^2 = \tfrac{1}{8}. $

Multiply through by $8$ to clear denominators:

$ 6p - 8p^2 = 1 \;\;\Longrightarrow\;\; 8p^2 - 6p + 1 = 0. $

Using the quadratic formula,

$ p = \frac{6 \pm \sqrt{36 - 32}}{16} = \frac{6 \pm \sqrt{4}}{16} = \frac{6 \pm 2}{16}. $

Thus,

$p = \tfrac{6 + 2}{16} = \tfrac{8}{16} = \tfrac{1}{2},$

$p = \tfrac{6 - 2}{16} = \tfrac{4}{16} = \tfrac{1}{4}.$

Since $p + q = \tfrac{3}{4}$,

if $p = \tfrac{1}{2}$, then $q = \tfrac{3}{4} - \tfrac{1}{2} = \tfrac{1}{4}$;

if $p = \tfrac{1}{4}$, then $q = \tfrac{3}{4} - \tfrac{1}{4} = \tfrac{1}{2}$.

So the two solutions are $\bigl(p,q\bigr) = \bigl(\tfrac{1}{2},\tfrac{1}{4}\bigr)$ or $\bigl(\tfrac{1}{4},\tfrac{1}{2}\bigr)$.

That is, the individual probabilities of $A$ and $B$ must be $\frac{1}{2}$ and $\frac{1}{4}$ (in either order).

Step 3: Match with the given options

The options are:

(A) $\tfrac{3}{8}$ and $\tfrac{1}{8}$

(B) $\tfrac{5}{8}$ and $\tfrac{1}{4}$

(D) $\tfrac{1}{2}$ and $\tfrac{1}{4}$

Clearly, Option D is exactly the pair $\bigl(\tfrac12,\tfrac14\bigr)$. (And it does not matter which event is which, since the question just asks for their individual probabilities.)

Hence, the correct answer is

$ \boxed{\text{Option D: } \frac{1}{2} \text{ and } \frac{1}{4}.} $

Probability

Continue this paper

Similar Probability PYQs

More practice from this question