What_is_the_probability_of_spinning_a_free_on_one_spin_and_an_s_on_your_second_if_you_have_a_spinner_with_1_space_labeled_free_and_the_remaining_11_mississippi_with_one_letter_in_each.So, you would multiply 1/12 times 4/12 to get 4/144 or 1/36
The chance of receiving a blue result is 2 in 4, in other words 50%.
1/4 raised to the power of 50; e.g. .2550.
4/6, or 66.67%.
Assuming you want the probability FOR A SINGLE TRY, and you want the numbers in that exact order, the probability for each part (for instance, first = red; or second = green) is 1/4; therefore, the probability for the combination is (1/4) to the power 4.
If there are four colors on a spinner, then the probability of spinning one particular color is 1 in 4, or 0.25. Also, the probability of spinning one of two particular colors is 2 in 4, or 0.5. Combining these two "unrelated" events simply requires multiplication. The probability, then, of spinning one particular color on one spin, and then spinning one of two particular colors on the next spin is (1 in 4) times (2 in 4), or 2 in 16, or 0.125.
The probability is 5/6. The die has 6 sides so 5 of them are not the "4". The probability of getting the four is 1/6.
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 spinning the number 3, or any number, is 1/4 or 0.25 since there is 4 numbers total.
What_is_the_probability_of_spinning_a_free_on_one_spin_and_an_s_on_your_second_if_you_have_a_spinner_with_1_space_labeled_free_and_the_remaining_11_mississippi_with_one_letter_in_each.So, you would multiply 1/12 times 4/12 to get 4/144 or 1/36
Just one head = 2/4 = 1/2
Assuming that the four-sided spinner is fair and that it is numbered in the traditional way of 1, 2, 3 and 4, the probability of spinning a three is 1/4.
The chance of receiving a blue result is 2 in 4, in other words 50%.
On one spin, and assuming the spinner is fair, 1/6
0.75%
4 of 4
1/4 raised to the power of 50; e.g. .2550.