A pair of 6 sided dice are tossed. What is the probability that at least one of the dice has a value greater than or equal to 4?

Question

We are working on probability in statistics right now and I can't seem to get the right answer on our homework for this question. 

Answer 1

And if you don't believe it,

1,1   1,2   1,3   1,4*  1,5*  1,6*

2,1   2,2   2,3   2,4*  2,5*  2,6*

3,1   3,2   3,3   3,4*  3,5*  3,6*

4,1* 4,2* 4,3*  4,4*  4,5*  4,6*

5,1* 5,2* 5,3*  5,4*  5,5*  5,6*

6,1* 6,2* 6,3*  6,4*  6,5*  6,6*

 

3/4 = 75% have a digit greater than or equal to 4

 

You can also visually see Elwyn's description.

1/2 of the rows, based on the first die (rows 4-6) are sucessful

for the other half, (rows 1-3), only half the time are they successful (die 3 = 4-6)

 

so 1/2 + 1/2of1/2 = 1/2 + 1/4 = 3/4

 

Answer 2

I would have to see your wrong answer to say why you went astray.

 

But the most common mistake on this type of question is due to counting rolls where both dice are 4 or better, twice.  If I roll a six and a four that constitutes one success.

 

The probability that the first die will be a four or better is 0.5.  So, half the time, the second die roll does not "matter," you already have a success.

 

So, half, of your rolls are successes even before the second die comes to a stop.  

 

The second half of you rolls will be 4 or better half the time.

 

P = 0.5  (for the first roll) + (0.5)(0.5) (half successes on the remaining half of the rolls) = 0.50 + 0.25 = 0.75

 

Xxxxxxx

 

The other method, and this is a powerful technique, is to calculate the chance of (double) failure, because P + ∼P = 1.0

 

The chance for a die roll to fail (∼P) is   ∼P = 0.5.

 

for both to fail, since they are independent events, is (0.5)(0.5)

 

So ,the probability of "at least one" is One minus the probability of neither.

1 - (0.5)(0.5) = 1 - 0.25 = 0.75

 

There are many problems where 1 - failure is much easier to calculate than success.