Math, asked by riyakushwaha348, 6 months ago

the different timing at which Sachin comes to the office on different days of the week is 1pm,3pm,8am, 4 am ,7 am,12pm, 5pm. find the probability that Sachin comes on a particular day before 10am.​

Answers

Answered by swayambhuvmitra83
0

Answer:

Let me see if I can make some other solution, by considering the discrete case.

Assume the arrivals happen only on the minutes, 0, 1, 2, 3 . . . 59, 60 (between 1PM and 2PM)

Number of cases of arrivals 61*61 = 3721.

If A arrives at 0, B can arrive at 0-15 (16 cases)

If A arrives at 1, B can arrive at 0-16 (17 cases)

If A arrives at 2, B can arrive at 0-17 (18 cases)

If A arrives at 3, B can arrive at 0-18 (19 cases)

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If A arrive at 15, B can arrive at 0-30 (31 cases)

If A arrive at 16, B can arrive at 1-31 (31 cases)

If A arrive at 17, B can arrive at 2-32 (31 cases)

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If A arrives at 44, B can arrive at 29-59 (31 cases)

If A arrives at 45, B can arrive at 30-60 (31 cases)

If A arrives at 46, B can arrive at 31-60 (30 cases)

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If A arrives at 56, B can arrive at 41-60 (20 cases)

If A arrives at 57, B can arrive at 42-60 (19 cases)

If A arrives at 58, B can arrive at 43-60 (18 cases)

If A arrives at 59, B can arrive at 44-60 (17 cases)

If A arrives at 60, B can arrive at 45-60 (16 cases)

Hence total success cases will be 16+17+18+19....30 (15 numbers) + 31+31 (31 numbers) + 30+29+....17+16(15 numbers) which is 345+961+345=1651

Probability=1651/3721=0.444

Answer by or = 7/16= 0.438.

When we can consider smaller discrete intervals of time (eg arrivals can occur on the second, 00:00,00:00:01,... 00:00:57,00:00:58,00:00:59,00:01:00,... 00:59:57,00:59:58,00:59:59,01:00:00), we get more closer to the exact answer.

When taking the limit (as time between possible arrivals tending to 0), we will get the same answer 7/16.

Step-by-step explanation:

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