In an experiment of rolling a regular six-sided dice once, what is the probability of getting a three given that the number is prime?
The key word in this question is the word given. We’re being asked to find a conditional probability, the probability of getting a three when we roll this dice given that the number we get is a prime number. This can be expressed using the notation of a vertical line to indicate the probability of three given that the number is prime. Now there are two ways that we can answer this question. The first way is to apply the conditional probability formula, which tells us that the probability of an event 𝐴 occurring given that another event 𝐵 has occurred is equal to the probability of the intersection of 𝐴 and 𝐵, that’s 𝐴 and 𝐵 both occurring, divided by the probability of event 𝐵 occurring.
In our question, event 𝐴 is the event that the number is a three. And event 𝐵 is the event that the number is prime. So we have that the probability of getting a three given that the number is prime is equal to the probability of getting a three and it being prime divided by the probability of the number being prime. Now, this is slightly unnecessary because three is a prime number and it’s the only prime number which is equal to three. So the probability of getting a number which is both three and prime is actually just equal to the probability of getting a three.
We’re told that this is a six-sided dice. So the numbers on its faces, the possible outcomes, are the integer values from one to six. And as this dice is regular, this means that the probability of the dice landing on each face, and therefore the probability of each outcome, is equal. We can therefore just count up the number of satisfactory outcomes in order to find the probabilities we’re looking for. There is just one three on the dice, and there are six outcomes in total. So the probability of getting a three is equal to one-sixth. There are three prime numbers on the dice, the numbers two, three, and five. So the probability of getting a prime number is three out of six or three-sixths. Remember, one is not a prime number because the definition of a prime number is that it has exactly two factors. And one has only one factor, which is just one.
Notice that three is the only number which has been circled twice. It’s the only number that’s in the intersection of the events three and prime. Which is why we said that the probability of getting a number which is both three and prime is in fact just equal to the probability of getting three. So substituting these probabilities into our conditional probability formula gives one-sixth over three-sixths. This, of course, means one-sixth divided by three-sixths. And we recall that in order to divide by a fraction, we flip or invert that fraction and multiply. So one-sixth divided by three-sixths is equal to one-sixth multiplied by six over three. We can cancel the six in the denominator of the first fraction with the six in the numerator of the second, giving just one-third for the value of this product.
So we found that the probability of getting a three given that the number is prime is one-third. Now let’s look at the second method, which is to use a Venn diagram. Normally, when we draw a Venn diagram, we would draw two overlapping circles. But, in this instance, the event of getting a three is a subset of the event of getting a prime number. So we draw the circle for three within the circle for prime numbers. We can then add all of the outcomes from one to six onto our Venn diagram. One is neither prime nor three. So it goes on the outside. Two is prime but not equal to three. So it goes inside the pink circle but not inside the orange circle. Three is both three and prime. So it goes inside the orange circle which is inside the pink circle.
Four is neither prime nor three, so it goes on the outside. Five is prime but not three. So it goes in the pink circle but not in the orange circle. And six is neither prime nor three. We then use our Venn diagram to find the required probability. As we know that the number is prime, the denominator for our fraction, the number of outcomes we’re interested in, is three because there are three numbers inside the pink circle. There’s only one number which is inside the orange circle for the event of three. So the numerator for our fraction is one, giving a probability of one-third.
Using two different methods then, we’ve shown that the probability of getting a three given that the number is prime in one roll of a regular six-sided dice is one-third.