p yes broke = .18
p not broke = .82
P(0 broke) = 1*.18^11 * 1
P(1 broke) = C(11,1).18^10 .82^1
P(2 broke) =C(11,2).18^9 .82^2
C(11,1) = 11
C(11,2) = 55
Industry standards suggest that 18% of new vehicles require warranty service within the first year. Jones Nissan, sold 11 Nissans yesterday
What is the probability that less than three of these vehicles require warranty service?
2 answers
sorry
P(0 broke) = 1* 1 * .82^11
P(1 broke) = C(11,1).18^1 .82^10
P(2 broke) =C(11,2).18^2 .82^9
P(0 broke) = 1* 1 * .82^11
P(1 broke) = C(11,1).18^1 .82^10
P(2 broke) =C(11,2).18^2 .82^9
1 answer