A company that makes cartons finds that the probability of producing a carton wi
ID: 3152413 • Letter: A
Question
A company that makes cartons finds that the probability of producing a carton with a puncture is .06, the probability that a carton has a smashed corner is .08, and the probability that a carton has a puncture and has a smashed corner is .005.
Answer parts (a) and (b) below.
(a) Are the events "selecting a carton with a puncture" and "selecting a carton with a smashed corner" mutually exclusive? Explain.
A. Yes, a carton can have a puncture and a smashed corner.
B. No, a carton cannot have a puncture and a smashed corner.
C. Yes, a carton cannot have a puncture and a smashed corner.
D. No, a carton can have a puncture and a smashed corner.
(b) If a quality inspector randomly selects a carton, find the probability that the carton has a puncture or has a smashed corner.
The probability that a carton has a puncture or a smashed corner is
.135
I know the first part is D, and that the probability is .135. What I don't know isi how to calculate the probability. Any help would be appreciated, thanks!
Explanation / Answer
A company that makes cartons finds that the probability of producing a carton with a puncture is .06, the probability that a carton has a smashed corner is .08, and the probability that a carton has a puncture and has a smashed corner is .005.
Answer parts (a) and (b) below.
(a) Are the events "selecting a carton with a puncture" and "selecting a carton with a smashed corner" mutually exclusive? Explain.
A. Yes, a carton can have a puncture and a smashed corner.
B. No, a carton cannot have a puncture and a smashed corner.
C. Yes, a carton cannot have a puncture and a smashed corner.
D. No, a carton can have a puncture and a smashed corner.
(b) If a quality inspector randomly selects a carton, find the probability that the carton has a puncture or has a smashed corner.
The probability that a carton has a puncture or a smashed corner is
.135
I know the first part is D, and that the probability is .135. What I don't know isi how to calculate the probability. Any help would be appreciated, thanks!
Addition rule is used. P(A or B)=P(A)+P(B)P(A and B)
P(P)=0.06
P(S) =0.08
P( P AND S)=0.005
P( P or S)= P( P)+P(S) - P( P AND S)
=0.06 + 0.08 - 0.005
=0.135
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