CBSE Sample Paper 2024-25 · Section C · 3 marks

Q31: Rain today and tomorrow, or $P(X < 3)$ when $P(X = r) \propto 5^{-r}$

Sample 2024-25Ch 13 · Probability

The question

(a) The probability that it rains today is 0.4. If it rains today, the probability that it will rain tomorrow is 0.8. If it does not rain today, the probability that it will rain tomorrow is 0.7. If

$P_1$ ⁠: denotes the probability that it does not rain today.

$P_2$ ⁠: denotes the probability that it will not rain tomorrow, if it rains today.

$P_3$ ⁠: denotes the probability that it will rain tomorrow, if it does not rain today.

$P_4$ ⁠: denotes the probability that it will not rain tomorrow, if it does not rain today.

(i) Find the value of $P_1 \times P_4 - P_2 \times P_3$ ⁠. (2 marks)

(ii) Calculate the probability of raining tomorrow. (1 mark)

(b) A random variable $X$ can take all non-negative integral values and the probability that $X$ takes the value $r$ is proportional to $5^{-r}$ ⁠. Find $P(X < 3)$ ⁠.

Part (b) uses a random variable and its probability distribution. This topic is no longer in the CBSE syllabus (2025-26).

Exam answer

(a)

Raining today and not raining today are complementary events, so

P_1 = 1 - 0.4 = 0.6

Why?

For any event $E$ ⁠, $P(\text{not } E) = 1 - P(E)$ ⁠, because exactly one of the two happens.

If it rains today, the probability that it will not rain tomorrow is

P_2 = 1 - 0.8 = 0.2

Why?

The complement rule also works for conditional probabilities with the same condition. Given rain today, "rain tomorrow" and "no rain tomorrow" are complementary, so their probabilities add to 1.

If it does not rain today,

P_3 = 0.7,\quad P_4 = 1 - 0.7 = 0.3

Why?

$P_3$ is given directly in the question. $P_4$ has the same condition (no rain today), so it is the complement of $P_3$ ⁠.

(i)

\begin{aligned} &P_1 \times P_4 - P_2 \times P_3 \\ &= 0.6 \times 0.3 - 0.2 \times 0.7 \\ &= 0.18 - 0.14 = 0.04 \end{aligned}

Why?

$6 \times 3 = 18$ and $2 \times 7 = 14$ ⁠. Each product of two one-decimal numbers has two decimal places.

\begin{aligned} 0.6 \times 0.3 &= 0.18 \\ 0.2 \times 0.7 &= 0.14 \\ 0.18 - 0.14 &= 0.04 \end{aligned}

(ii) Let $E_1$ and $E_2$ be the events that it will rain today and it will not rain today respectively, and $E$ the event that it will rain tomorrow.

P(E_1) = 0.4,\quad P(E_2) = 0.6

P(E \mid E_1) = 0.8

P(E \mid E_2) = 0.7

Why?

$E_1$ and $E_2$ cannot both happen, and one of them must happen. So they split all the possibilities into two cases, which is what the theorem of total probability needs.

The numbers come from the question and from the values found above: $P(E_2) = P_1 = 0.6$ ⁠, and $P(E \mid E_2) = P_3 = 0.7$ ⁠.

By the theorem of total probability,

\begin{aligned} P(E) &= P(E_1)P(E \mid E_1) \\ &\quad + P(E_2)P(E \mid E_2) \\ &= 0.4 \times 0.8 + 0.6 \times 0.7 \\ &= 0.32 + 0.42 = 0.74 \end{aligned}

Why?

Rain tomorrow can happen after rain today or after a dry day. Add the two routes, each weighted by the chance of its condition.

$0.4 \times 0.8 = 0.32$ and $0.6 \times 0.7 = 0.42$ ⁠.

\begin{aligned} 4 \times 8 &= 32 \Rightarrow 0.32 \\ 6 \times 7 &= 42 \Rightarrow 0.42 \\ 0.32 + 0.42 &= 0.74 \end{aligned}

The probability of rain tomorrow is 0.74.

(b)

Given $P(X = r) \propto \dfrac{1}{5^r}$ ⁠, so

P(X = r) = \frac{k}{5^r}\quad (r = 0, 1, 2, \ldots)

where $k$ is a non-zero constant.

Why?

"Proportional to" means "equal to a constant times". Also $5^{-r} = \frac{1}{5^r}$ ⁠.

\begin{aligned} &P(X = 0) + P(X = 1) \\ &\quad + P(X = 2) + \ldots = 1 \end{aligned}

Why?

In a probability distribution, the probabilities of all the values add up to 1. Here $X$ can be any non-negative integer, so the sum has infinitely many terms.

\Rightarrow k\left(1 + \frac15 + \frac{1}{5^2} + \frac{1}{5^3} + \ldots\right) = 1

\Rightarrow k\left(\frac{1}{1 - \frac15}\right) = 1 \Rightarrow k = \frac45

Why?

The bracket is an infinite geometric series with first term 1 and common ratio $\frac15$ ⁠. Since $\left|\frac15\right| < 1$ ⁠, its sum is

\frac{a}{1 - r} = \frac{1}{1 - \frac15} = \frac54

\begin{aligned} 1 - \frac15 &= \frac55 - \frac15 = \frac45 \\ \frac{1}{\frac45} &= 1 \times \frac54 = \frac54 \end{aligned}

So $\frac54k = 1$ ⁠, which gives $k = \frac45$ ⁠.

\begin{aligned} \frac54k &= 1 \\ k &= 1 \times \frac45 = \frac45 \quad (\times\,\tfrac45) \end{aligned}

\begin{aligned} &P(X < 3) \\ &= P(X = 0) + P(X = 1) \\ &\quad + P(X = 2) \\ &= \frac45\left(1 + \frac15 + \frac{1}{5^2}\right) \\ &= \frac45\cdot\frac{31}{25} \end{aligned}

Why?

$X$ takes only the values 0, 1, 2, ..., so $X < 3$ means $X = 0$ ⁠, 1 or 2.

Over the common denominator 25:

1 + \frac15 + \frac{1}{25} = \frac{25 + 5 + 1}{25}

( $1 = \frac{25}{25}$ ⁠, $\frac15 = \frac{5}{25}$ ⁠, $5^2 = 25$ ⁠.)

\begin{aligned} 25 + 5 + 1 &= 31 \\[4pt] \frac45 \cdot \frac{31}{25} &= \frac{4 \times 31}{5 \times 25} \end{aligned}

Then $4 \times 31 = 124$ and $5 \times 25 = 125$ ⁠.

$\therefore\ P(X < 3) = \dfrac{124}{125}$ ⁠.

Where this comes from

Chapter 13, Probability: Total probability; probability distribution.

Question type: (a)(ii) Two routes to one result

How to spot it: click to see

(a)(ii) 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: Rain tomorrow can follow rain today (0.4, then 0.8) or no rain today (0.6, then 0.7). Multiply along each route and add: $0.32 + 0.42 = 0.74$ ⁠.

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Q31: Rain today and tomorrow, or P(X<3)P(X < 3)P(X<3) when P(X=r)∝5−rP(X = r) \propto 5^{-r}P(X=r)∝5−r

Q31: Rain today and tomorrow, or $P(X < 3)$ when $P(X = r) \propto 5^{-r}$