Question

How can i find the probability of obtaining certain results?(very urgent for anupcoming test, details inside)

4 dices are thrown (the dices are regular):

a)if 4 even(not odd) results were obtained, what are the chances all of them are larger than 3?

b)if we get at least 2 even(not odd) results, what are the chances that between the 4 results there is at least one result that is equal to 6?

it is very important for an upcoming test, please help me if you can

Answer 1

See below:

Explanation:

a

On the throw of 4 dice with results that are all even, we have 3 possible results: 2, 4, 6. The probability that they are all greater than 4 can be found by using a binomial probability.

`sum_(k=0)^(n)C_(n,k)(p)^k(1-p)^(n-k)=1`

We have `n=4` (the four dice).

The probability that any given roll is greater than 3 is `2/3`, and so `p=2/3`.

We're looking at `k=4` - the event where all the dice are successfully above 3. This all gives:

`((4),(4))(2/3)^4(1/3)^0~~0.1975`

b

Let's think about this question in this way - while the conditions will be met if either the known even rolls or the other rolls achieve a 6, the only way to not the meet the condition of achieving a 6 is if both sets of dice don't roll one. And so if we calculate the probability for the known even dice to not roll a 6 and we multiply by the other roll also not achieving a 6, we'll have the probability of not rolling a 6 with any of the four dice. We can then subtract that value from 1.

Known evens (2 dice)

We have `n=2, p=1/3` and we want `k=0`

`((2),(0))(1/3)^0(2/3)^2=4/9`

The other two dice

We have `n=2, p=1/6, k=0`

`((2),(0))(1/6)^0(5/6)^2=25/36`

Putting it together

The probability of not having a 6 on any of the four dice is:

`4/9xx25/36=25/81~~0.3086`

Meaning that the probability of having at least one 6 is:

`1-0.3086~~0.6914`