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What is the probability of a number less than 6 on the number cube followed by a number equal to 6 on the spinner?
The probability is(5 times the number of 6s on the spinner/6 timesthe total number of different positions on the spinner)
Suppose the arrow is spun 50 times about how many times would you expect the spinner to lamd on an odd number?
You can expect the spinner to land an odd number 25 times out of 50.
You spin a spinner two times find the probability that the spinner stops on 3 then 1?
The probability that a spinner with N sides stops on 2 particular numbers in two spins in 1 in N2. It does not matter what the two numbers are, since the two spins are sequentially unrelated.
The spinner is divided into 12 equal parts if you spin the spinner 100 times how many times do you expect the spinner to stop on a number less than 10?
∙It will be spun on a number less than 10 in 75 times if you spin it 100 times.Explanation:Let x be the random variable, x = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.The probability for any outcome is P(x) = 1/12. It is a flat probability distribution.The probability that when you spin the spinner the outcome is a number greater than10 is:P(10 ≤ x ≤ 12) = P(x = 10) + P(x = 11) + P(x = 12 ) = 3/12 = 1/4The probability of the complement event (that the outcome is a number smaller than 10) is: P(x < 10) = 1 - 1/4 = 3/4.So the expected number of times the spinner outcome will be a number smaller than 10 is:P(No. outcomes x
If you spin this spinner 36 times how many times would you expect it to land on 2?
The answer depends on the number of sides on the spinner and how they are numbered.
What is the probability of a number less than 6 on the number cube followed by a number equal to 6 on the spinner?
The probability is(5 times the number of 6s on the spinner/6 timesthe total number of different positions on the spinner)
Suppose the arrow is spun 50 times about how many times would you expect the spinner to lamd on an odd number?
You can expect the spinner to land an odd number 25 times out of 50.
suppose the arrow is spun 50 times. about how many times would you expect the spinner to land on an odd number the spinner is divided by 5?
5
During an experiment, a spinner landed on red 6 times. If the resulting experimental probability of the spinner landing on red is StartFraction 1 over 8 EndFraction, how many trials were performed?
Given, The probability of getting red, P(R) = 1/8 Red occurs by the spinner= 6 times Let, the total number of trials = N Therefore, for the experimental probability the total number of trials performed can be calculated by the following equation: P(R) = (Red occurs by the spinner)/(Total number of trials) Or, 1/8 = 6/N Or, N = 6 × 8 Or, N = 48 Final Answer: A spinner landed on red 6 times. If the resulting experimental probability of the spinner landing on red is StartFraction 1 over 8 EndFraction, then 48 trials were performed.
You spin a spinner two times find the probability that the spinner stops on 3 then 1?
The probability that a spinner with N sides stops on 2 particular numbers in two spins in 1 in N2. It does not matter what the two numbers are, since the two spins are sequentially unrelated.
If you spin the spinner two times what is the probability that the spinner will land on the black region twice?
To calculate the probability of spinning the black region twice on a spinner, you first need to determine the total number of possible outcomes when spinning the spinner twice. Let's say the spinner has 8 equal sections, with 2 black regions. The total outcomes for spinning the spinner twice would be 8 x 8 = 64. The probability of landing on the black region twice would be 2/8 x 2/8 = 4/64 = 1/16. Therefore, the probability of landing on the black region twice is 1/16 or approximately 0.0625.
The spinner is divided into 12 equal parts if you spin the spinner 100 times how many times do you expect the spinner to stop on a number less than 10?
∙It will be spun on a number less than 10 in 75 times if you spin it 100 times.Explanation:Let x be the random variable, x = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.The probability for any outcome is P(x) = 1/12. It is a flat probability distribution.The probability that when you spin the spinner the outcome is a number greater than10 is:P(10 ≤ x ≤ 12) = P(x = 10) + P(x = 11) + P(x = 12 ) = 3/12 = 1/4The probability of the complement event (that the outcome is a number smaller than 10) is: P(x < 10) = 1 - 1/4 = 3/4.So the expected number of times the spinner outcome will be a number smaller than 10 is:P(No. outcomes x
What is the probability that the arrow will land on either Saturday or Sunday both times?
The answer depends on where the arrow is being thrown!
If you spin this spinner 36 times how many times would you expect it to land on 2?
The answer depends on the number of sides on the spinner and how they are numbered.
What is the probability that you will spin a 4 and roll a 4?
Assuming each possible number on a spinner has the same probability and an unbiased die is being rolled, the answer depends on how many numbers are on the spinner, and how many times the number 4 appears on each.To find the probability, workout the probability of spinning a 4 on the spinner and the probability of rolling a 4 on the die; then as spinning the spinner has no effect on rolling the die, they are independent events and to get the probability of both happening multiply them together.The probability of success is the number of successful outcomes divided by the total number of outcomes, giving:Probability(spinning a 4) = how_many_4s_are_on_the_spinner / how_many_numbers_are_on_the_spinnerProbability(rolling a 4) = how_many_4s_are_on_the_die / how_many_numbers_are_on_the_dieProbability(spinning a 4 and rolling a 4) = Probability(spinning a 4) × Probability(rolling a 4)Examples:an octagonal spinner with the numbers 1-4 on it each twice and a tetrahedral die (as used in D&D games) with the numbers 1-4 on it→ pr(spin 4 & roll 4) = 2/8 × 1/4 = 1/16a decagonal spinner with the numbers 0-9 and a tetrahedral die with the numbers 0-3 on it→ pr(spin 4 & roll 4) = 1/10 × 0/4 = 0a decagonal spinner with the numbers 0-9 and a standard die with the numbers 1-6 on it→ pr(spin 4 & roll 4) = 1/10 × 1/6 =1/60
If you spin the spinner 600 times how many times would you expect it to land on green?
Well, if the spinner has equal sections, and green is one of them, then statistically speaking, you would expect it to land on green about 100 times out of 600 spins. But hey, life's full of surprises, so don't bet your retirement savings on it!
What is the probability of spinning a 2 on a 1-6 spinner if I spin it 60 times?
The probability of getting a 2 is 1 - (1/6)60 = 1 - 2.05*10-47
