Probability: Question 4

Worked solution

Part (a): Both detectors raise the alarm

Since the two detectors act independently, the “and” of both events uses the multiplication rule: $P(D_1 \text{ and } D_2) = P(D_1) \times P(D_2) = 0.9 \times 0.8 = 0.72$

Part (b): Neither detector raises the alarm

The complementary probabilities that each detector does not raise the alarm are $1-0.9=0.1$ and $1-0.8=0.2$. Because the detectors are independent, so are their complements: $P(\text{neither}) = (1-0.9)(1-0.8) = 0.1 \times 0.2 = 0.02$

Part (c): At least one detector raises the alarm

“At least one” is the complement of “neither”: $P(\text{at least one}) = 1 - P(\text{neither}) = 1 - 0.02 = 0.98$

Check by the addition rule: using $P(D_1 \text{ or } D_2) = P(D_1) + P(D_2) - P(D_1 \text{ and } D_2)$, $P(\text{at least one}) = 0.9 + 0.8 - 0.72 = 1.7 - 0.72 = 0.98$ which agrees with the complement method.

Part (d): Exactly one detector raises the alarm

“Exactly one” splits into two mutually exclusive cases, “Detector 1 fires and Detector 2 doesn’t” or “Detector 2 fires and Detector 1 doesn’t”, so their probabilities are added: $P(\text{exactly one}) = P(D_1)(1-P(D_2)) + (1-P(D_1))P(D_2) = (0.9\times0.2) + (0.1\times0.8) = 0.18+0.08=0.26$

Check by another method: “exactly one” is “at least one” with the “both” case removed: $P(\text{exactly one}) = P(\text{at least one}) - P(\text{both}) = 0.98 - 0.72 = 0.26$ which agrees exactly.

Overall consistency check: the three mutually exclusive outcomes “both”, “exactly one” and “neither” must account for every possibility, so their probabilities should sum to $1$: $P(\text{both}) + P(\text{exactly one}) + P(\text{neither}) = 0.72 + 0.26 + 0.02 = 1.00 \checkmark$

Final answers

• (a) $P(\text{both}) = \boxed{0.72}$
• (b) $P(\text{neither}) = \boxed{0.02}$
• (c) $P(\text{at least one}) = \boxed{0.98}$
• (d) $P(\text{exactly one}) = \boxed{0.26}$