Exercise 13.2 · Q7: Find p: mutually exclusive, then independent

Basics you need

Part (i): $A$ and $B$ mutually exclusive

1. What "mutually exclusive" gives

   Mutually exclusive means $A\cap B = \phi$⁠, so $P(A\cap B) = 0$⁠, and the addition rule becomes

   $$P(A\cup B) = P(A) + P(B).$$
   With no overlap there is nothing counted twice, so the correction term $P(A\cap B)$ is zero and "or" is just a sum.
   Addition rule for mutually exclusive events (Class XI). Basics: the addition rule

2. Substitute and solve for $p$

   $$\begin{aligned} \frac{3}{5} &= \frac{1}{2} + p \\ p &= \frac{3}{5} - \frac{1}{2} \\ &= \frac{6 - 5}{10} = \frac{1}{10}. \end{aligned}$$
   Subtract $\frac12$ from both sides. The common denominator of 5 and 2 is 10: $\frac35 = \frac{6}{10}$ and $\frac12 = \frac{5}{10}$⁠.
   The addition rule of Part (i), Step 1, with $P(A) = \frac12$⁠, $P(B) = p$ and $P(A\cup B) = \frac35$ put in. Basics: the addition rule

3. Check

   $P(A) + P(B) = \frac12 + \frac{1}{10} = \frac{6}{10} = \frac35 = P(A\cup B)$⁠. Also $p = \frac{1}{10}$ is between 0 and 1, and $P(A) + P(B) = \frac35 \le 1$⁠, as it must be for two events that cannot overlap.

   Substituting back confirms the arithmetic, and the range check confirms such events can exist.
   Every probability lies between 0 and 1 (Class XI). Basics: Venn diagrams

Part (ii): $A$ and $B$ independent

1. What "independent" gives

   $$\begin{aligned} P(A\cap B) &= P(A)\,P(B) \\ &= \frac{1}{2}\,p = \frac{p}{2}. \end{aligned}$$
   Independence turns the unknown overlap into a product of known quantities, so the addition rule again has only one unknown, $p$⁠.
   Definition 3, Section 13.4. Lesson: independent events

2. Substitute into the addition rule and solve

   Put $P(A\cup B) = \frac35$⁠, $P(A) = \frac12$⁠, $P(B) = p$ and $P(A\cap B) = \frac{p}{2}$ into $P(A\cup B)$ $= P(A) + P(B) - P(A\cap B)$⁠:

   $$\begin{aligned} \frac{3}{5} &= \frac{1}{2} + p - \frac{p}{2} \\ \frac{3}{5} &= \frac{1}{2} + \frac{p}{2} \\ \frac{p}{2} &= \frac{3}{5} - \frac{1}{2} = \frac{1}{10} \\ p &= \frac{2}{10} = \frac{1}{5}. \end{aligned}$$
   Collect the $p$ terms: $p - \frac{p}{2} = \frac{p}{2}$⁠. Then subtract $\frac12$ and multiply both sides by 2.
   Addition rule (Class XI), with the overlap from Part (ii), Step 1. Basics: the addition rule

3. Check

   With $p = \frac15$⁠: $P(A\cap B) = \frac12\times\frac15 = \frac{1}{10}$⁠, and

   $$\begin{aligned} P(A\cup B) &= \frac{1}{2} + \frac{1}{5} - \frac{1}{10} \\ &= \frac{5 + 2 - 1}{10} = \frac{6}{10} = \frac{3}{5}.\;\checkmark \end{aligned}$$
   The overlap equals $P(A)\,P(B)$⁠, so the events really are independent with this $p$⁠, and the union comes out as given.
   Mutually exclusive and independent are different (Remarks (ii), Section 13.4). Basics: Venn diagrams

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