In a hotel, rooms are numbered from 101 to 550. A room is chosen at random. What is the probability that room number starts with 1, 2 or 3 and ends with 4, 5 or 6 ?
Out of which 299 room number starts with either 1, 2 or 3. (as room number 100 is not there) Now out of those 299 rooms only 90 room numbers end with 4, 5 or 6
So the probability is 90/450 i.e. 1/5 or 0.20
Total favorable ways 3c1*10c1*3c1 i.e. 3*10*3 = 90 So probability = 90/450 or 9/45
3c1*10c1*3c1
3*10*3 = 90
The probability is 90/450 i.e. 1/5 or 0.20
Where 450 are total number of room and 90 rooms starts with 1, 2 or 3 and ends with 4, 5 or 6
as rooms started from 101 to 550, it means there are 5 floors and each floor has 50 rooms. total rooms are 250. no. of rooms started with 1,2 &3 and ends with 4,5,6 are 75. thus, (75/250)*100=30% so, probabilty is 30%
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