That would depend on how many numbers are on the spinner and the cube. The more numbers there are, the less likely it is that they would both land an any given number.
The answer depends on how many numbers are 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)
Because the number cube is not sentient enough to know the result of the spinner and modify its own outcome accordingly. And conversely, the outcome of the spinner is not affected by the roll of the cube.
For an ordinary number cube, the answer is 1/6
The probability of rolling a seven with one roll of a standard number cube is zero.
The answer depends on how many numbers are on the spinner.
The answer will depend on how many numbers are on the spinner.
The answer depends on how many sides the spinner has.
3/6=1/2 on a number cube it depends on how many sides of a spinner there are if its an even number always 1/2 if its an odd number of sides it cant be simplified
The probability is(5 times the number of 6s on the spinner/6 timesthe total number of different positions on the spinner)
The answer depends on WHAT is landed: a number cube, a tetrahedral die, some other polyhedron, a spinner?
There is insufficient information on the shape of the spinner.
Presuming that the spinner and the number cube are both "fair", then no - spinning the spinner and tossing the six-sided number cube are called statistically independent events. They do not influence each other, and it does not matter which order the events occur in.
Because the number cube is not sentient enough to know the result of the spinner and modify its own outcome accordingly. And conversely, the outcome of the spinner is not affected by the roll of the cube.
Pr(Sum > 25) = Pr(Spinner = 30 or 40 and Cube = 6) = Pr(Spinner = 30 or 40)*Pr(Cube = 6) = 2/4 * 1/6 = 1/2*1/6 = 1/12 or 8.33... %
0.5 in each case. The probability of both happening simultaneously is 1/4
independent