pr(six) = 1/6 → expected 6s in 90 rolls = 1/6 × 90 = 15
The expected value is 20 times.
fifty percent
The expected number of rolls under 3 is the number of rolls times the probability of rolling less than 3. So: E(rolls less than 3 out of 30) = 30 * 1/3 = 30/3 = 10
100/6 = 16.6 times
pr(six) = 1/6 → expected 6s in 90 rolls = 1/6 × 90 = 15
Three times.
The expected value is 20 times.
The expected number of times is the probability x number of throws. Since you have a prob of 1/6 for a seven, then (1/6) * 160 = 26.67 times you would have success. We generally would round up the expected number of successes to the whole number 27.
fifty percent
The expected number of rolls under 3 is the number of rolls times the probability of rolling less than 3. So: E(rolls less than 3 out of 30) = 30 * 1/3 = 30/3 = 10
100/6 = 16.6 times
The expectation is 50 times.
If a DIE (not dice) is rolled 90 times, the expected value of the sum of the first and second rolls is 7 if you assume that the die is fair. It does not matter how many times you roll the die, as long as it is at least 2.
Type your answer here... 15
Never. If you roll it 1 time, there can be no "then".
16.66... % of the time - if the die is fair. And the percentage is not affected by the number of times you roll it.