CBSE Sample Paper 2020-21 · Part A: Section I · 1 mark

Q16: Probability that it rains only on the first 4 days of the week

Sample 2020-21Ch 13 · Probability

The question

The probability that it will rain on any particular day is 50%. Find the probability that it rains only on first 4 days of the week.

Exam answer

P(rain on a day) $= \frac{1}{2}$ and P(no rain on a day) $= 1 - \frac{1}{2} = \frac{1}{2}$ ⁠; the days are independent.

Why?

50% means $\frac{50}{100} = \frac{1}{2}$ ⁠. "No rain" is the complement of "rain", so its probability is $1 - \frac{1}{2}$ ⁠.

\begin{aligned} \frac{50}{100} &= \frac{50 \div 50}{100 \div 50} = \frac{1}{2} \\ 1 - \frac{1}{2} &= \frac{2}{2} - \frac{1}{2} = \frac{1}{2} \end{aligned}

It must rain on each of the first 4 days and not rain on the other 3 days of the week.

Why?

"Only on the first 4 days" means two things: rain on days 1, 2, 3 and 4, and no rain on days 5, 6 and 7. A week has 7 days, so all 7 days are fixed.

\begin{aligned} P &= \left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^3 \\ &= \left(\frac{1}{2}\right)^7 \end{aligned}

Why?

For independent days, the probability that all 7 things happen is the product of the 7 probabilities. Add the powers: $4 + 3 = 7$ ⁠, and $2^7 = 128$ ⁠.

\begin{aligned} P &= \left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^3 \\ &= \left(\frac{1}{2}\right)^{4 + 3} = \left(\frac{1}{2}\right)^7 \\ &= \frac{1}{2^7} = \frac{1}{128} \end{aligned}

Double 2 until there are seven 2s multiplied: $2, 4, 8, 16, 32, 64, 128$ ⁠. So $2^7 = 128$ ⁠.

Watch out: $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$ is the probability of rain on the first 4 days with no condition on the other 3. It ignores the word "only".

$\therefore$ Required probability $= \dfrac{1}{128}$ ⁠.

Where this comes from

Chapter 13, Probability: Independent events: rain on given days.

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Probability in past papers: question 36 of 38 · see them all