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31 votes
31 votes

$A$ 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. $9/16$
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8 Answers

Best answer
48 votes
48 votes

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: $C$

edited by
27 votes
27 votes

Check this

9 votes
9 votes
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
6 votes
6 votes

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.

Answer:

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