Sample 2023-24 Class 12 Maths Q14 Solution: Problem solved by at least one of three students (chances 1/2, 1/3, 1/4)

The question

A problem in Mathematics is given to three students whose chances of solving it are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ respectively. If the events of their solving the problem are independent then the probability that the problem will be solved, is

(a) $\frac{1}{4}$

(b) $\frac{1}{3}$

(c) $\frac{1}{2}$

(d) $\frac{3}{4}$

Exam answer

Let $E_1, E_2, E_3$ be the events that the three students solve the problem. They are independent, so $\bar E_1, \bar E_2, \bar E_3$ are also independent.

Why?

When events are independent, their complements are independent too. So the probabilities of "not solving" can be multiplied.

\begin{aligned} P(\bar E_1) &= 1 - \frac{1}{2} = \frac{1}{2} \\ P(\bar E_2) &= 1 - \frac{1}{3} = \frac{2}{3} \\ P(\bar E_3) &= 1 - \frac{1}{4} = \frac{3}{4} \end{aligned}

The problem is solved if at least one of them solves it.

Required probability $= 1 - P(\bar E_1)P(\bar E_2)P(\bar E_3)$

Why?

"At least one solves it" is the complement of "nobody solves it", so

\begin{aligned} &P(\text{solved}) \\ &= 1 - P(\text{none solves}) \end{aligned}

Nobody solves it when $\bar E_1$ ⁠, $\bar E_2$ and $\bar E_3$ all happen.

\begin{aligned} &= 1 - \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \\ &= 1 - \frac{1}{4} = \frac{3}{4} \end{aligned}

Why?

The product cancels neatly:

\begin{aligned} \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} &= \frac{1 \times 2 \times 3}{2 \times 3 \times 4} \\ &= \frac{6}{24} \\ &= \frac{6 \div 6}{24 \div 6} = \frac{1}{4} \end{aligned}

Then write 1 as quarters to subtract:

\begin{aligned} 1 - \frac{1}{4} &= \frac{4}{4} - \frac{1}{4} \\ &= \frac{4 - 1}{4} = \frac{3}{4} \end{aligned}

Watch out: adding $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{13}{12}$ is wrong. It is more than 1, because two students can both solve it.

Answer: (d) $\dfrac{3}{4}$

Where this comes from

Chapter 13, Probability: Independent events: probability of at least one.

Question type: At least one, or exactly one

How to spot it: click to see

Type 6. At least one, or exactly one: think "not" first

Spot it: independent events, with "at least one", "the problem is solved", "neither" or "exactly one".

Here: Three students work independently, and "the problem will be solved" means at least one solves it. So take 1 minus the chance all three fail: $1 - \frac12\cdot\frac23\cdot\frac34 = \frac34$ ⁠.

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Q14: Problem solved by at least one of three students (chances 12,13,14\frac{1}{2}, \frac{1}{3}, \frac{1}{4}21​,31​,41​⁠)

Q14: Problem solved by at least one of three students (chances $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ ⁠)