The probability that out of four residential fires who independently reported on a single day, two are in family homes, one is in an apartment, and one is in another type of dwelling is 0.0895 or 8.95%.
The multinomial distribution deals specifically with events that have multiple discrete outcomes. The multinomial distribution is not limited to events with only discrete outcomes.
We have given that
The National Fire Incident Reporting Service stated, among all residential fires.
Let consider the number of events ,
X₁---> fires residential who are in home
X₂--> fires residential who are in appartment
X₃--> fires residential who are in another type of dwelling
The probability of ae X₁ occured (p₁)
= 73% = 0.73
The probability of event X₂ occured (p₂) = 20% =0.20
The probability of event X₃ occurred (p₃) = 7% = 0.07
Four residential fires are independently reported on a single day.
we have to calculate probability that two are in family homes, one is in an apartment, and one is in another type of dwelling?
Now, Using the Multinomial distribution,
X₁= 2 , X₂= 1, X₃ = 1 and n = 4
P( X₁, X₂,-----,Xₙ ) =( n!/(X₁! X₂! ----Xₙ!) )( p₁ˣ₁ × p₂ˣ₂ × ----× pₙˣₙ)
P( 2,1,1) = 4!/2! 1! 1! ( 0.73² × 0.2× 0.07¹)
= 12 ( 0.014 × 0.5329)
= 0.0895272
Hence, the required probability is 0.089..
To learn more about Multinomial distribution, refer:
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Complete question:
The National Fire Incident Reporting Service stated that, among residential fires, 73% are in family homes, 20% are in apartments, and 7% are in other types of dwellings.
a) If four residential fires are independently reported on a single day, what is the probability that two are in family homes, one is in an apartment, and one is in another type of dwelling?
