Exercise 13.2 · Q14: A and B solve a problem independently

Basics you need

Part (i): the problem is solved

1. Name the events

   Let $E$ be the event "A solves the problem" and $F$ the event "B solves the problem". Then

   $$\begin{aligned} P(E) &= \frac{1}{2}, & P(E') &= 1 - \frac{1}{2} = \frac{1}{2}, \\ P(F) &= \frac{1}{3}, & P(F') &= 1 - \frac{1}{3} = \frac{2}{3}. \end{aligned}$$

   $E$ and $F$ are independent, because A and B work independently.

   The complements $E'$ ("A fails") and $F'$ ("B fails") will be needed in both parts, so we find them now with $P(E') = 1 - P(E)$⁠.
   $P(E') = 1 - P(E)$⁠. Basics: events as sets

2. Use the complement of "neither solves it"

   The problem is solved when at least one of them solves it, the event $E\cup F$⁠. It is not solved only when both fail, the event $E'\cap F'$⁠. Since $E'$ and $F'$ are independent,

   $$\begin{aligned} P(E\cup F) &= 1 - P(E'\cap F') \\ &= 1 - P(E')\,P(F') \\ &= 1 - \frac{1}{2}\times\frac{2}{3} \\ &= 1 - \frac{1}{3} = \frac{2}{3}. \end{aligned}$$
   "At least one solves it" has three cases (A only, B only, both); its complement "neither" is one case, and for independent events it is a product. The Note after Example 13 is what allows $P(E'\cap F') = P(E')\,P(F')$⁠; Example 14 is exactly this formula.
   Example 14 and the Note after Example 13, Section 13.4. Lesson: Examples 10 to 14

3. Second route: the addition rule

   $$\begin{aligned} &P(E\cup F) \\ &= P(E) + P(F) - P(E)\,P(F) \\ &= \frac{1}{2} + \frac{1}{3} - \frac{1}{6} \\ &= \frac{3 + 2 - 1}{6} = \frac{2}{3}. \end{aligned}$$
   This route uses only the independence of $E$ and $F$ themselves: $P(E\cap F) = \frac12\times\frac13 = \frac16$⁠. Getting $\frac23$ twice confirms the answer.
   Addition rule (Class XI) and Definition 3, Section 13.4. Basics: the addition rule

Part (ii): exactly one of them solves it

1. Split into the two cases

   Exactly one solves it when A solves and B fails ($E\cap F'$⁠), or B solves and A fails ($E'\cap F$⁠). These cannot both happen, so

   $$\begin{aligned} &P(\text{exactly one}) \\ &\quad = P(E\cap F') + P(E'\cap F). \end{aligned}$$
   In the first case A solves it and in the second A does not, so no outcome is in both cases. Probabilities of mutually exclusive events add.
   Mutually exclusive events: probabilities add. Basics: Venn diagrams

2. Each case is a product

   $$\begin{aligned} P(E\cap F') &= P(E)\,P(F') \\ &= \frac{1}{2}\times\frac{2}{3} = \frac{1}{3}, \\ P(E'\cap F) &= P(E')\,P(F) \\ &= \frac{1}{2}\times\frac{1}{3} = \frac{1}{6}. \end{aligned}$$

   So

   $$\begin{aligned} P(\text{exactly one}) &= \frac{1}{3} + \frac{1}{6} \\ &= \frac{2 + 1}{6} = \frac{1}{2}. \end{aligned}$$
   $E$ and $F'$ are independent (Example 13), and so are $E'$ and $F$ (the Note after it), which allows each product. The grid shows the same split: "A only" is 2 cells and "B only" is 1 cell out of 6.
   Example 13 and the Note after it, Section 13.4. Lesson: independent events

3. Check: "solved" minus "both"

   "Exactly one" is "at least one" without "both":

   $$\begin{aligned} P(E\cup F) - P(E\cap F) &= \frac{2}{3} - \frac{1}{6} \\ &= \frac{4 - 1}{6} = \frac{1}{2}. \end{aligned}$$
   The event "both solve it" lies inside "at least one solves it"; removing it leaves exactly the two cases of Part (ii), Step 1. The same $\frac12$ again.
   Two events cut $S$ into four regions. Basics: Venn diagrams

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