The probability of throwing any two numbers on a die - say, either a 1 or a 2 - on a single throw is one chance out of three, or 33%.
When calculating the probability of repeating favorable throws with a single die the probability of throwing a 1 or a 2 twice in succession is 1/9, which is the square of one chance out of three, or 1/3 multiplied by itself. Therefore, the probability of throwing a 1 or 2 three times in succession would be 1/27, or 1/3 x 1/3 x 1/3 , while the probability of throwing a 1 or a 2 four times in succession would be 1/3 to the fourth power.
We now want to apply probability theory logic to the probability of throwing a 1 or a 2 with a pair of dice, instead of with a single die. If the probability of throwing a 1 or a 2 with a single die is one out of three we can assume that throwing a 1 or a 2 with two dice would be twice as great, or 67%. Is this a correct assumption? Please explain your reasoning.© BrainMass Inc. brainmass.com June 3, 2020, 9:50 pm ad1c9bdddf
If we want to throw two dices and getting 1 or 2 . In this case our sample space has ...
The solution calculates the probability of obtaining two numbers on a die.