Question

small question
...what is the answer for (iv)
i already answer other three parts of the question :
14a=1-(0.10+0.20)
so, a= 0.7/14 = 0.05

(X) P(X)
0 0.10
1 0.20
2 0.45
3 0.15
4 0.05
5 0.05

(i) 0.15
(ii) (0.20 + 0.45 + 0.15 + 0.05) = 0.85
(iii) (0.15 + 0.05 + 0.05) = 0.25
(iv) ?????????????? i need a help...

Answer 1

the question in attached picture ... please help ...

Answer 2

Well, by simply looking at the table, what do you make out about the probability for more than 5 accidents a day ?

Answer 3

there is no accidents for more than 5 days included in the table
...what is the correct reasoning?

Answer 4

Well then the probability is simply 0. ^_^
SInce the sum of probabilities for the mentioned events is already 1, there won't be any chance of anything happening outside the given domain/range.
I am not too sure how to express this correctly.

Answer 5

wow .. don't worry friend
thank you very much...

Answer 6

your probability function is valid on the support {0,1,2,3,4,5} and 0 otherwise. Since P(X>5) implies a support of x=6,7,8... then its just 0 as @shubhamsrg mentioned

Answer 7

nice brother kirbykirby.... you have help me before also.. thanks for all of you ... now i can write my correct answer..

Answer 8

yw