6.8k views

$A$ and $B$ are friends. They decide to meet between 1:00 pm and 2:00 pm on a given day. There is a condition that whoever arrives first will not wait for the other for more than $15$ minutes. The probability that they will meet on that day is

1. $1/4$
2. $1/16$
3. $7/16$
4. $9/16$

edited | 6.8k views

Meeting occurs if the first person arrives between $1:00$ and $1:45$ and the second person arrives in the next $15$ minutes or if both the persons arrive between $1:45$ and $2:00.$

Case 1:

• $45/60$ are favourable cases and hence probability of first person arriving between $1:00$ and $1:45$ is $3/4.$
• Probability of second person arriving in the next $15$ minutes $= 15/60 = 1/4$
• So, probability of one person arriving between $1:00$ and $1:45$ and meeting the other $= 3/4 \times 1/4 \times 2 = 3/8$ $(2$ for choosing the first arriving friend$)$

Case 2:

• Both friends must arrive between $1:45$ and $2:00.$ Probability $= 1/4 \times 1/4 = 1/16.$

So, probability of a meet $= 3/8 + 1/16 = 7/16$

Correct Answer: $C$

by Veteran (422k points)
edited
0
why not in second case u did 2C1?
+2
For the second case both have to arrive after 1:45- we do not have a choice here. In the first case, we have an order- first arriving can be one among 2.
0

@ sir,

But in 1:45 to 2:00

If A comes first then B Or B comes first then A

doen't this count to two diff cases ??

For such questions which are also known as probability based on areas , u can find easily using grid of 2 dimension as shown below

 1/2*F F 1/2 * F F 1/2 * F 1/2*F F 1/2 * F F 1/2 * F

The favorable area for the given problem is shown above...

So no of favourable cells  =  4  +  6 * 1/2     =    7

Therefore probability that they meet  =  No of favourable cells/ No of total cells

=   7 / 16

Hence C) should be the correct option.

by Veteran (101k points)
0
But you ignore the case when A reaches first..
0
For arrival of B we are considering vertical axis.
0
yes. But consider the first 'F' from origin. Actually, it won't be F, even if B comes after A. But you consider only B coming first.
0
0
+1
Corrected..:)
+5
Hi habib could you explain it bit more.

We are unable to understand what those cells in table are representing ..?
probability that one person meet on that day = 15/60 = 1/4
Prob(failing to meet) = 3/4
prob(failing to meet by both the persons) = 3/4 * 3/4 = 9/16
Prob( meet on that day by both the persons) = 1-9/16 = 7/16
by Active (1.4k points)

Check this

by Active (1.8k points)
+1
+1
Best one :)
+1 vote
Mark this formula for this kind of any question.

x=minutes one could not wait for.

n=60/x.

probability they meet= (2n-1)/n^2

here x=60/15 =4.

probabi;ity=8-1/16 =7/16
by (193 points)
7/16
by (27 points)
We can derive one formula also from graph For(n*n)

(n^2-(n-1)^2) /n^2

from given ques n=4

(16-9)/16

by Active (2.1k points)
0
Could you explain bit more please.