Question

A renowned hospital usually admits 200 patients every day. One per cent patients, on an
average, require special room facilities. On one particular morning, it was found that only
one special room is available. What is the probability that more than 3 patients would
require special room facilities?​

Answer 1

Answer:

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Answer 2

find : distribution of 3 patients requiring special room facility

given : 200 persona daily admitted in hospital and 1 percent of them ( 2 ) is need special facility

solution :

more than 3 person mean...3,4,5, ... 200

so expected value oposite 0,1,2

so p= 1- ( X = x [0,1,2] )

we get answer by Poisson distribution =

$e ^{ - m} .m ^{x } \div x \: fectorial$

notice take m = 2

= 1 - [ e"-m + e"-m×m¹/1! + e"-m ×m²/2! + e"-m×m³/3! ]

= 1 - e-² [ 1+ 2/1 + 4/2 + 8/6 ]

= 1 - e-² [ 1+2+2+4/3 ]

= 1 - e-² [ 5+4/3 ]

= 1 - e-² [ 19/3 ]

= 1 - ( 1/e² × 19/3 ) value of e = 2.71828

= 1 - ( 0.135335× 19/3 )

= 1 - ( 0.857121 )

= 0.1428

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