Question

two friends decides to meet at a location X between 4 to 5 pm. they decided that when someone comes they have to wait maximum 30 mins for the other to arrive if they are late for eq if someone comes at 4 they wait till 4:30 for the other and if still they didnt come the person will leave. what is the probability that they will meet

Answer 1

3/4

Answer 2

To determine the probability that the two friends will meet, we can use geometric probability.

Let $x$ be the arrival time of the first friend (in minutes past 4 pm) and $y$ be the arrival time of the second friend (in minutes past 4 pm). Since they decide to meet between 4 pm and 5 pm, the total time interval is 60 minutes. So, $0 \le x \le 60$ and $0 \le y \le 60$.

The sample space is a square region in the $xy$-plane with vertices (0,0), (60,0), (60,60), and (0,60). The area of this sample space is $60 \times 60 = 3600$ square units.

The friends will meet if the absolute difference between their arrival times is less than or equal to the maximum waiting time, which is 30 minutes. So, the condition for them to meet is $|x - y| \le 30$.

This inequality can be broken down into two parts:

1. $x - y \le 30 \implies y \ge x - 30$
2. $-(x - y) \le 30 \implies y - x \le 30 \implies y \le x + 30$

We need to find the area of the region within the square $0 \le x \le 60, 0 \le y \le 60$ that satisfies both $y \ge x - 30$ and $y \le x + 30$.

This region can be visualized by considering the total square and subtracting the "unfavorable" regions where they do not meet. The unfavorable regions are two triangles at the corners of the square.

1. Region where $y > x + 30$: This represents the case where the second friend arrives more than 30 minutes after the first friend, and the first friend leaves. This region is a triangle with vertices:

   • (0, 30) (intersection of $x=0$ and $y=x+30$)
   • (0, 60) (corner of the square)
   • (30, 60) (intersection of $y=60$ and $y=x+30 \implies 60 = x+30 \implies x=30$) The legs of this right-angled triangle are both 30 units (from $x=0$ to $x=30$ and from $y=30$ to $y=60$). Area of this triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 30 \times 30 = 450$ square units.

2. Region where $y < x - 30$: This represents the case where the first friend arrives more than 30 minutes after the second friend, and the second friend leaves. This region is a triangle with vertices:

   • (30, 0) (intersection of $y=0$ and $y=x-30 \implies 0 = x-30 \implies x=30$)
   • (60, 0) (corner of the square)
   • (60, 30) (intersection of $x=60$ and $y=x-30 \implies y=60-30 \implies y=30$) The legs of this right-angled triangle are both 30 units (from $x=30$ to $x=60$ and from $y=0$ to $y=30$). Area of this triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 30 \times 30 = 450$ square units.

The total unfavorable area (where they do not meet) is $450 + 450 = 900$ square units.

The favorable area (where they meet) is the total sample space area minus the unfavorable area: Favorable Area = $3600 - 900 = 2700$ square units.

The probability that they will meet is the ratio of the favorable area to the total sample space area: Probability = $\frac{\text{Favorable Area}}{\text{Total Area}} = \frac{2700}{3600} = \frac{27}{36} = \frac{3}{4}$.

The probability that they will meet is $\frac{3}{4}$.