The probability that when you toss a dye two times (or toss two dice) both times the
dye shows the same number (or the two dice roll doubles) is calculated as follows.
First dye toss roll any outcome (1, 2, 3, 4, 5 or 6). The probability the second toss
comes out the same as first toss is 1/6. So, for two dice to roll doubles:
P(doubles) = 1/6
Probability of complement event (not rolling doubles) is;
P(not doubles) 5/6
So the probability of first two dice trial 'not doubles' and following trial of the two dice
'doubles is;
P([N doubles]1 and [doubles]2) = 5/6 ∙ 1/6 = 5/36
The probability is based on the second cube having 6 sides. No matter what you roll on the first cube, the chance of rolling "doubles" is 1 in 6! Now if you ask, what is the chance in rolling a specific set of doubles, the chance is 1 in 6 times 1 in 6 times, or 1 in 36 ...
2/12 or 1/6
Assuming you are talking about fair, six-sided dice, then the probability of rolling a 1 on the first roll in 1/6, and the probability of rolling a 2 on the second roll is 1/6. Putting these together, the probability of rolling 1 on the first die and 2 on the second die is 1/36. If you do not care about the order, then you could roll 1,2 or 2,1; in this case the probability would be 2/36, or 1/18.
The question asks "What is the probability of rolling either an even number on the first roll or a 1 on the second roll?" These events are independent from each other as the outcome of the second roll is not affected by the outcome of the first roll. However, these events are non-mutually exclusive, meaning that these events can both occur at the same time.The probability of rolling an even number on the first roll is 3/6 because 2, 4, and 6 are even numbers and a six-sided die has six possible numbers.The probability of rolling a 1 on the second roll is 1/6.If these two probabilities are added together, we will have "double counted" the event where an even number is rolled on the first roll and a 1 is rolled on the second roll. To correct for this, we must subtract the probability of both events occurring.The probability that both events occur is 3/36, because 3/6 * 1/6 = 3/36.Now, the probability of rolling either an even number on the first roll or a 1 on the second roll is:3/6 + 1/6 - 3/36= 18/36 + 6/36 - 3/36= 21/36= 7/12
Rolling a sum of 15 on three rolls of a die, when the first roll is a 4, is the same as rolling a sum of 11 on the second and third roll. The probability of rolling 11 on two dice is 3 in 36, or 1 in 12.
The probability that you will roll doubles on a pair of dice is 1 in 6. The probability that you roll "something" on the first die is 1 in 1. The probability that the second die will match the first die is 1 in 6. The resultant probability is simply the product of (1 in 1) and (1 in 6).
1/6 on the first roll, and 1/6 of rolling a six on the second roll. Overall, the chances of you rolling a 5 on the first and a six on the second is 1/3.
The probability is based on the second cube having 6 sides. No matter what you roll on the first cube, the chance of rolling "doubles" is 1 in 6! Now if you ask, what is the chance in rolling a specific set of doubles, the chance is 1 in 6 times 1 in 6 times, or 1 in 36 ...
When you roll doubles in Monopoly, you get to take another turn. If you roll doubles three times in a row, you go to jail.
When you roll doubles in Monopoly, you get to take another turn. If you roll doubles three times in a row, you go to jail.
1/6 + 1/6 = 1/3
It is 1/12.
The probability of rolling a 2 on the first roll is 1 in 6. The probability of rolling a 3 on the second roll is 1 in 6. However, the probability of rolling a 2 on the first roll and 3 on the second roll before you roll at all is 1/6 x 1/6 = 1 in 36.
If you roll doubles in Monopoly, you get to take another turn.
1/6*1/2 = 1/12
and the question is ... ?
2/12 or 1/6