CBSE Sample Paper 2026-27 · Section D · 5 marks

Q35: Smartwatch false alerts: total probability and Bayes' theorem

Sample 2026-27Ch 13 · Probability

The question

A fitness app syncs with four types of smartwatches: Alpha, Beta, Gamma and Delta. 45% of the users use Alpha, 30% use Beta,15% use Gamma and 10% use Delta.

Due to device errors, the probabilities of getting a false heart-rate alert from the Alpha, Beta, Gamma and Delta smartwatches are respectively 0.01, 0.02, 0.05, and 0.01.

Based on the given information, answer the following questions:

(i) Find the probability that a randomly selected user gets a false alert.

(ii) If a false alert occurs, what is the probability that it came from a Beta smartwatch?

Exam answer

Let the events be

$E_1$ ⁠: user uses Alpha smartwatch, $E_2$ ⁠: user uses Beta smartwatch, $E_3$ ⁠: user uses Gamma smartwatch, $E_4$ ⁠: user uses Delta smartwatch, $A$ ⁠: user gets a false heart-rate alert.

$\{E_1, E_2, E_3, E_4\}$ is a partition of the sample space.

Why?

Each user uses exactly one of the four watches, so the four events do not overlap and together they cover every user. That is a partition. Check: the probabilities add up to 1.

\begin{aligned} &0.45 + 0.3 + 0.15 + 0.1 \\ &= 1 \end{aligned}

\begin{aligned} 0.45 + 0.3 &= 0.75 \\ 0.75 + 0.15 &= 0.9 \\ 0.9 + 0.1 &= 1 \end{aligned}

\begin{aligned} P(E_1) &= 0.45,\ P(E_2) = 0.3, \\ P(E_3) &= 0.15,\ P(E_4) = 0.1 \end{aligned}

\begin{aligned} P(A|E_1) &= 0.01,\ P(A|E_2) = 0.02 \\ P(A|E_3) &= 0.05,\ P(A|E_4) = 0.01 \end{aligned}

Why?

The percentages of users give $P(E_i)$ ⁠: 45% is 0.45, and so on.

The error rates are conditional probabilities. "The probability of a false alert from a Beta watch" means the chance of a false alert given that the user has a Beta watch, so it is $P(A|E_2)$ ⁠, not $P(A)$ ⁠.

(i) By the theorem of total probability,

Why?

A false alert can come through any of the four watches. The theorem of total probability adds the four routes:

P(A) = \sum_{i = 1}^{4} P(E_i)P(A|E_i)

\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &\quad + P(E_3)P(A|E_3) \\ &\quad + P(E_4)P(A|E_4) \\ &= 0.45 \times 0.01 + 0.3 \times 0.02 \\ &\quad + 0.15 \times 0.05 + 0.1 \times 0.01 \\ &= 0.0045 + 0.006 \\ &\quad + 0.0075 + 0.001 \\ &= 0.019 \end{aligned}

Why?

Multiply each pair, then add:

\begin{aligned} 0.45 \times 0.01 &= 0.0045 \\ 0.3 \times 0.02 &= 0.006 \\ 0.15 \times 0.05 &= 0.0075 \\ 0.1 \times 0.01 &= 0.001 \end{aligned}

Line up the decimal places when adding:

\begin{aligned} &0.0045 + 0.0060 \\ &+ 0.0075 + 0.0010 \\ &= 0.0190 \end{aligned}

Two at a time:

\begin{aligned} 0.0045 + 0.0060 &= 0.0105 \\ 0.0105 + 0.0075 &= 0.0180 \\ 0.0180 + 0.0010 &= 0.0190 \end{aligned}

$\therefore$ Probability of a false alert $= 0.019$

(ii) By Bayes' theorem,

Why?

Now the alert has happened, and we ask which watch it came from. That reverse question is Bayes' theorem: the Beta route's share of the total.

P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(A)}

The bottom is the total $P(A)$ from part (i).

\begin{aligned} P(E_2|A) &= \frac{P(E_2)P(A|E_2)}{P(A)} \\ &= \frac{0.3 \times 0.02}{0.019} = \frac{0.006}{0.019} \end{aligned}

Why?

Multiply the top and the bottom by 1000 to clear the decimals: $\frac{0.006}{0.019} = \frac{6}{19}$ ⁠.

\begin{aligned} 0.3 \times 0.02 &= 0.006 \\[4pt] \frac{0.006}{0.019} &= \frac{0.006 \times 1000}{0.019 \times 1000} \\ &= \frac{6}{19} \end{aligned}

6 and 19 have no common factor, so $\frac{6}{19}$ is in its lowest terms.

$\therefore$ Required probability $= \dfrac{6}{19}$

Where this comes from

Chapter 13, Probability: Total probability and Bayes' theorem.

Question type: (i) Two routes to one result; (ii) Bayes' theorem

How to spot it: click to see

(i) Type 7. Two routes to one result: total probability

Spot it: the result can happen in two (or more) ways, such as a ball from bag I or from bag II, and you want its overall chance.

Here: A false alert can come from any of the four watches (45%, 30%, 15%, 10%). Multiply each by its false-alert rate and add: 0.019.

(ii) Type 8. Bayes' theorem: from the result back to the cause

Spot it: the result has already happened (a red ball, a faulty item, an A grade), and you are asked for the chance that it came from one cause. It often comes as a case study or a 5-mark question.

Here: A false alert has already happened, and you want the chance it came from a Beta watch. Divide the Beta route, $0.3 \times 0.02 = 0.006$ ⁠, by the total 0.019 from (i).

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