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Full Question
<em>Of the cartons produced by a company, 10% have a puncture, 6% have a smashed corner, and 0.4% have both a puncture and a smashed corner. Find the probability that a randomly selected carton has a puncture or a smashed corner. The probability that a randomly selected carton has a puncture or a smashed corner nothing ____%. (Type an integer or a decimal. Do not round.)</em>
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Answer:
Step-by-step explanation:
Given
Required
For non-mutually exclusive event described above, P(Punctured or Smashed Corner) can be calculated as thus;
Substitute:
10% for P(Puncture Corner),
6% for P(Smashed Corner) and
0.4% for P(Punctured and Smashed Corner)
Convert % to fraction
Convert to decimal
I hope my answer help you in your question
Answer :
<h3>Step-by-step explanation:
Given that a multiple-choice test contains 10 questions and there are 4 possible answers for each question.
∴ Answers=4 options for each question.
<h3>To find how many ways a student can answer the given questions on the test if the student answers every question :</h3>Solving this by product rule
Product rule :
<u>If one event can occur in m ways and a second event occur in n ways, the number of ways of two events can occur in sequence is then m.n</u>
From the given the event of choosing the answer of each question having 4 options is given by
The 1st event of picking the answer of the 1st question=4 ,
2nd event of picking the answer of the 2nd question=4 ,
3rd event of picking the answer of the 3rd question=4
,....,
10th event of picking the answer of the 10th question=4.
It can be written as by using the product rule