Q.The probability of finding a green signal on a busy crossing X is 30%. What is the probability of finding a green signal on X on two consecutive days out of three?

Concept: Independent events — the multiplication theorem. Each day's signal is independent of the others, so the probability of a particular sequence of days is the product of the daily probabilities.

Let $p = P(\text{green}) = 0.3$ and $q = P(\text{not green}) = 1 - 0.3 = 0.7$ on any single day.

We need the signal green on two consecutive days out of the three. Over days 1-2-3 the only patterns with two adjacent green days (and the third day not green) are:

• GGR — green on days 1 and 2
• RGG — green on days 2 and 3

(The pattern GRG has greens on days 1 and 3, which are not consecutive, so it is excluded.)

By independence: …

Why this approach works

The signal on each day is an independent event with a fixed probability, so we use the multiplication theorem for independent events: the probability of a whole day-by-day sequence is the product of the individual daily probabilities. Then, because different sequences are mutually exclusive, we add the probabilities of the sequences that satisfy the condition.

The crucial word in the question is consecutive. We are not asked for "green on any two of the three days" — we are asked for green on two days that are next to each other. That distinction changes which day-patterns count as favourable.

Step-by-step solution

1. Set up the per-day probabilities.

   • $P(\text{green on a day}) = p = 0.3$
   • $P(\text{not green on a day}) = q = 1 - p = 0.7$ The three days are independent.

2. List all patterns of "green on exactly two days" over days 1-2-3.

   Writing G = green and R = not green, exactly two greens can occur as:

3. Keep only the patterns where the two green days are consecutive.

   • GGR — greens on days 1 and 2, consecutive (favourable)
   • RGG — greens on days 2 and 3, consecutive (favourable)
   • GRG — greens on days 1 and 3, not consecutive (day 2 sits between them), so excluded

   So only GGR and RGG are favourable. …

Method: Probability of a specified pattern over independent repeated days

When a question asks for a particular arrangement of successes across independent trials, list exactly the favourable arrangements, price each by multiplying independent per-day probabilities, then add.

Steps

Step 1: Fix the per-trial probabilities.

Each day is independent with success probability $p$ (green) and failure $q=1-p$.

Step 2: Read the wording precisely and list the favourable patterns. …