Answer :
Answer:
66.667%
Step-by-step explanation:
In a fair die, there are 6 possible rolls. Out of the 6, 1 number is chosen each time. Since order does not matter in this problem, we will be applying combination, so this would be written as 6C1 x 4. In order to exclude 3 and 4, from 6, the choices must be narrowed down to 4. Written out as an equation, this would be 4C1 x 4. To get the percentage, simply divide the two equations.
(4C1 x 4)/(6C1 x 4) = 0.66667 = 66.667%
We want to find the joint probability of rolling a fair die 4 times and not getting any 3 nor 4 in these rolls.
We will see that the probability is:
P = 16/81 = 0.198
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Let's see how to solve this:
In a fair die, all the possible outcomes (numbers of the set {1, 2, 3, 4, 5, 6}) have the same probability, which is 1/6 (one over the total number of outcomes)
So the probability of getting a given outcome from that set is 1 (this means that we always get an outcome from that set).
Now if we remove two of these values (the 3 and the 4).
The probability will be:
p = 1 - 1/6 - 1/6 = 4/6 = 2/3.
And the fair die is rolled 4 times, then in each roll, we have a probability of 4/6 of not getting a 3 nor a 4.
The joint probability (the probability of not getting a 3 nor a 4 in any of these rolls) is equal to the product between the individual probabilities, so we will get:
P = p^4 = (2/3)*(2/3)*(2/3)*(2/3) = 16/81 = 0.198
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