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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?
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Answer / manpreet singh
Answer is 1/5(0.20)
Total rooms=550-101+1=450
Favourable rooms= 1z4(where z=0,1,2,3,4,5,6,7,8,9)=10
1z5(where z=0,1,2,3,4,5,6,7,8,9)=10
1z6(where z=0,1,2,3,4,5,6,7,8,9)=10
2z4(where z=0,1,2,3,4,5,6,7,8,9)=10
2z5(where z=0,1,2,3,4,5,6,7,8,9)=10
2z6(where z=0,1,2,3,4,5,6,7,8,9)=10
3z4(where z=0,1,2,3,4,5,6,7,8,9)=10
3z5(where z=0,1,2,3,4,5,6,7,8,9)=10
3z6(where z=0,1,2,3,4,5,6,7,8,9)=10
total favourable rooms=90
probability=90/450=1/5=0.20
| Is This Answer Correct ? | 19 Yes | 2 No |
Answer / guest
There are total 450 rooms.
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
| Is This Answer Correct ? | 12 Yes | 2 No |
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