Question

What is the chance of throwing at least 7 in a single cast, with two dice?
(a) $\dfrac{5}{12}$
(b) $\dfrac{7}{12}$
(c) $\dfrac{1}{4}$
(d) $\dfrac{17}{36}$

Answer 1

Hint: We must list all the possible outcomes for this experiment, and count the total number of outcomes. We must then list all those outcomes in which the sum is greater than or equal to 7, and count them to find the number of favourable outcomes. We can then take the ratio of number of favourable outcomes and total number of outcomes to get the probability.

Complete step by step answer:
We know that a chance or a probability is defined as the ratio of number of favourable outcomes and total number of outcomes.
Chance or probability = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
Here, in this problem, we have two dice which are thrown in a single cast. We are then adding the numbers on the top face of each dice to get an outcome for the experiment.
Let us now list all the possible numbers to appear on the two dice. So, here are all the possible outcomes,
{(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)}
Here, we can see that the total number of outcomes is 36.
Now, we need at least 7. So, the sum of the numbers on the dice must be greater than or equal to 7.
Let us now list all favourable outcomes,
{(1,6),
(2,5), (2,6),
(3,4), (3,5), (3,6),
(4,3), (4,4), (4,5), (4,6),
(5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Here, we can clearly see that the total number of favourable outcomes is 21.
Thus, we can now write
Probability = $\dfrac{\text{21}}{\text{36}}$
We can cancel the common factors from both numerator and denominator, to get
Probability = $\dfrac{7}{12}$
Thus, the chance of throwing at least 7, in a single throw, using two dice is $\dfrac{\text{7}}{\text{12}}$ .

So, the correct answer is “Option b”.

Note: We must read the question very carefully and understand that the two dice are different, and so (1,2) and (2,1) are two different outcomes, and we must not consider them as similar to one another. We must always keep in mind that the value of probability is always between 0 and 1.