Suppose two friends Anjali and Prabhat trying to meet for a date to have lunch say between 2 pm to 3 pm. Suppose they follow the following rules for this meeting:

Each of them will arrive either on time or 12 minutes late or 24 minutes late or 36 minutes late or 48 minutes late or 1 hour late. All these arrival times are equally likely for both of them.

Whoever of them reaches first will wait for the other to meet only for 10 minutes. If within 10 minutes the other does not reach, he/she leaves the place and they will not meet.

Find the probability of their meeting.

(b) In the study learning material (SLM), you have seen many situations where Poisson distribution is suitable and discussed some examples of such situations. Create your own example for a situation other than those that are discussed in SLM. If you denote your created random variable by X then find the probability that X is less than 2.

(a)
Let us define the possible arrival times (in minutes after 2:00 PM) as the set:
{0,12,24,36,48,60}\{0, 12, 24, 36, 48, 60\}
Each time is equally likely. So total possible arrival combinations of (Anjali, Prabhat) = 6×6=366 \times 6 = 36.

Let Anjali arrive at time AA and Prabhat at time PP. They meet if:
∣A−P∣≤10|A - P| \leq 10 (i.e., the second person arrives within 10 minutes of the first)

We now count how many pairs (A,P)(A, P) satisfy this condition:

• (0,0)(0, 0) → yes

• (12,12)(12, 12) → yes

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