It depends on how many points there are that the spinner can land on. If there are 8, for example, the probability would be 8/16, or 1/2...
The probability is(5 times the number of 6s on the spinner/6 timesthe total number of different positions on the spinner)
You can expect the spinner to land an odd number 25 times out of 50.
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.
∙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
The answer depends on the number of sides on the spinner and how they are numbered.
The probability is(5 times the number of 6s on the spinner/6 timesthe total number of different positions on the spinner)
You can expect the spinner to land an odd number 25 times out of 50.
5
4 of 4
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.
∙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
The answer depends on where the arrow is being thrown!
The answer depends on the number of sides on the spinner and how they are numbered.
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
The probability of getting a 2 is 1 - (1/6)60 = 1 - 2.05*10-47
3/5=g/30
The answer will depend on what the experiment is: rolling a die, spinning a spinner, the number of times someone will lose before they win (or the converse), the number of rooms in a house, or whatever. Since you have not bothered to share that crucial bit of information, I cannot provide a more useful answer.