Respuesta :
Answer:
[tex] P(X \geq 3)= 1- P(X<3)= 1-P(X \leq 2)= 1- [P(X=0) +P(X=1) +P(X=2)][/tex]
And we can find the individual probabilites using the probability mass function and we got:
[tex] P(X=0) = (15C0) (0.08)^{0} (1-0.08)^{15-0}=0.286 [/tex]
[tex] P(X=1) = (15C1) (0.08)^{1} (1-0.08)^{15-1}=0.373 [/tex]
[tex] P(X=2) = (15C2) (0.08)^{2} (1-0.08)^{15-2}=0.227 [/tex]
And replacing we got:
[tex] P(X\geq 3) = 1-[0.286+0.373+0.227 ]= 0.114[/tex]
Step-by-step explanation:
For this case we can assume that the variable of interest is "drivers were involved in a car accident last year" and for this case we can model this variable with this distribution:
[tex] X \sim Bin (n =15, p =0.08)[/tex]
And for this case we want to find this probability;
[tex] P(X \geq 3)[/tex]
and we can use the complement rule and we got:
[tex] P(X \geq 3)= 1- P(X<3)= 1-P(X \leq 2)= 1- [P(X=0) +P(X=1) +P(X=2)][/tex]
And we can find the individual probabilites using the probability mass function and we got:
[tex] P(X=0) = (15C0) (0.08)^{0} (1-0.08)^{15-0}=0.286 [/tex]
[tex] P(X=1) = (15C1) (0.08)^{1} (1-0.08)^{15-1}=0.373 [/tex]
[tex] P(X=2) = (15C2) (0.08)^{2} (1-0.08)^{15-2}=0.227 [/tex]
And replacing we got:
[tex] P(X\geq 3) = 1-[0.286+0.373+0.227 ]= 0.114[/tex]
