Calculate the probability of getting a number that is either even or greater than 3

Having trouble wrapping my head around this. The question asks "Calculate the probability of getting a number that is either even or greater than 3"

Because this an OR question, my thinking was:

The probability of rolling an even number is: 3/6 (2,4,6)
The probability of rolling a number greater than 3 is 3/6 (4,5,6)

So the probability that your roll is either going to be an even number OR a number greater than 3 is 1/2.

At least that’s my logic (which obviously is incorrect.)

Can I get any help?

so let’s look at the sample space for a single die
{1, 2, 3, 4, 5, 6}

Remember we have or not and if the condition was the probability of getting a number that is greater than 3 and even then it is 1/2

  1. A number greater than 3 gives a sample space of {4, 5, 6}
  2. Since we have the or condition meaning that the condition is satisfied with the above sample space we have 1/3

Can you elaborate more how did you get \frac{1}{2}, here? because (\frac{3}{6} + \frac{3}{6}) would be 1.

This is still wrong probability as they explained in next mission that addition rule that you have learned in previous mission does not work for events that share the output.

The further explanation about it are in these next mission

I am sure you will get clear about it after taking these mission. I hope it helps.

Hi @39230239.jakaka,

Firstly, let us digress a bit to recap what an event is and how we calculate its probability. For eg: in tossing two coins, the sample space is S = {HH,HT,TH,TT}, where H = Heads and T = Tails. Any valid subset of S, say M = {HH,HT}, can be termed an event. M is said to have occurred if an element of M is obtained when the experiment is carried out. To calculate the probability of an event, we consider all the possible outcomes that could lead to that event occurring, count these (here, M has 2 elements), and divide it by the total number of outcomes in the sample space (4). So, P(M) = 1/2.

As you mentioned, the problem statement is to find the probability of getting a number that is either even or greater than 3, when we roll a fair die. We can consider this as two separate events, A, B defined as,

  • Rolling an even number, A = {a: a is an even number and a ϵ S} = {2,4,6}
  • Rolling a number greater than 3, B = {b: b > 3 and b ϵ S} = {4,5,6}.
  • where, Sample space of rolling a die, S = {1,2,3,4,5,6}

Now, let us define a new event, E as {e: (e is an even number) or (e is greater than 3), e ϵ S}. By intuition, we have E = {2,4,5,6}. We can represent all this in the Venn diagram as below:
Event A is the smaller, red circle; B is the larger green circle and in the general case, there could be an overlap between the two (yellow region). S is the light blue region. So, as E = {2,4,5,6}, we have P(E) = number of outcomes in E/number of elements in S = 4/6 = 2/3.
Another simple way is to use the formula for the probability of composite events, as below:
P(E) = P(AUB) = P(A) + P(B) - P(AB)
Here, U => Union of sets, ꓵ => Intersection of sets.
Simply put, the above equation translates to:
probability of A or B = probability of event A + probability of event B - (probability of the overlap between A, B). We can calculate each P() as, P(A) = 3/6; P(B) = 3/6; P(AB) = 2/6, and find P(E) = P(AUB) = 2/3, again.

Hope this helps :slight_smile: