The monthly worldwide average number of airplane crashes of commercial airlines

varaderiyw

varaderiyw

Answered question

2021-11-18

There are 3.5 commercial airliner aircraft crashes on average each month around the world. What is the likelihood that (a) there will be at least two of these accidents in the upcoming month; (b) there will only be one accident in the upcoming month? Describe your thinking!

Answer & Explanation

Phisecome

Phisecome

Beginner2021-11-19Added 18 answers

An average of 3.5 aircraft crashes occur per month. We must determine the likelihood that at least two accidents of this nature will occur in the upcoming month. Let n represent the number of monthly flights. Of course n is big. Also let Xi be a Bernoulli random variable where P(Xi=1)=p represents a probability that a plane will crash. Obviously
X1+X2++XnB(n,p)
The binomial distribution is approximated by a Poisson distribution since n is very high (the number of flights is supposed to be large) and we may assume that the likelihood of crash is very tiny (it is evident). Consequently, a random variable
S=X1+X2++XnP(λ)
where λ=np=E[S]. We need to calculate P(S2). Let's remember that P(X=k)=ϵλk! if XP(λ).
It follows:
P(S2)=1P(S=0)P(S=1)
=1e3.53.5e3.5
=10.033.50.03=0.86
In the (b) part we are calculating the porbability of at most 1 accident happening. Similar as before we have:
P(S1)=P(S=0)+P(S=1)=e3.5+3.5e3.5=0.136
We calculate the probability for the (a) part, which is 0.86 and for the (b) part, which is 0.136.

Sarythe

Sarythe

Beginner2021-11-20Added 11 answers

Step-by-step explanation:
We would apply the formula for poisson distribution which is expressed as
P(x=r)=(eμ×μr)r!
Where
μ represents the mean of the theoretical distribution.
r represents the number of successes of the event.
From the information given,
μ=3.5
a)
For the probability that there will be at least 2 such accidents in the next month, it is expressed as
P(x2)=1P(x<2)
P(x<2)=P(x=0)+P(x=1)
Therefore,
P(x=0)=(e3.5×3.50)0!
P(x=0)=(e3.5×1)1
P(x=0)=0.03
P(x=1)=(e3.5×3.51)1!
P(x=1)=(e3.5×3.5)1
P(x=1)=0.11
P(x<2)=P(x=0)+P(x=1)=0.03+0.11=0.14
P(x2)=10.14=0.86
b)
For the probability that there will be at most 1 accident in the next month, it is expressed as
P(x1)=P(x=0)+P(x=1)
(x=0)=0.03
(x=1)=0.11
P(x1)=0.03+0.11=0.14

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