If a quiz has 20 multiple choice questions, then the expected number of correct answers is 5.
For each question, there are 4 possible choices, and only 1 of them is correct.
So, the probability of randomly guessing the "correct-answer" for any question is = 1/4 = 0.25,
Each question is an independent event, we use the concept of linearity of expectation to calculate the expected number of correct answers for all 20 questions.
The "Expected" number of correct answers for a "single-question" is probability of guessing it correctly, which is 0.25,
The "Expected-Number" of correct answers for 20 questions is the sum of the expected number of correct answers for each question,
⇒ Expected number of correct answers = (Probability of guessing a single question correctly) × (Number of questions),
Substituting the values,
We get,
⇒ Expected number of correct answers = 0.25 × 20,
⇒ Expected number of correct answers = 5
Therefore, the expected number of correct answers when making "random-guesses" on all 20 multiple choice questions is 5.
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Consider a sequence whose first five terms are:-1.75, -0.5, 0.75, 2, 3.25
Which explicit function (with domain all integers n ≥ 1) could be used to define and continue this sequence?
Step-by-step explanation:
+ 1.25
every new term is the previous term + 1.25.
with starting value -1.75
f(n) = 1.25n - 1.75
of the cartons produced by a company, 3% have a puncture, 6% have a smashed corner, and 1.4% 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 is 0.076, or 7.6%.
What is probability?
Probability is a measure of the likelihood of an event occurring. It is a number between 0 and 1, where 0 means the event is impossible and 1 means the event is certain to happen.
To find the probability that a randomly selected carton has a puncture or a smashed corner, we can use the formula:
P(puncture or smashed corner) = P(puncture) + P(smashed corner) - P(puncture and smashed corner)
where P(puncture) is the probability of a carton having a puncture, P(smashed corner) is the probability of a carton having a smashed corner, and P(puncture and smashed corner) is the probability of a carton having both a puncture and a smashed corner.
Substituting the given probabilities into the formula, we get:
P(puncture or smashed corner) = 0.03 + 0.06 - 0.014
P(puncture or smashed corner) = 0.076
Therefore, the probability that a randomly selected carton has a puncture or a smashed corner is 0.076, or 7.6%.
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