Question

In a simultaneous throw of a pair of dice, find the probability of getting:
(a) 8 as the sum.
(b) A doublet.
(c) A doublet of prime numbers.
(d) A doublet of odd numbers.
(e) A sum greater than 9.

Answer 1

Hint: Start by drawing the table of all the possible outcomes. There would be a total of 36 outcomes, as 6 options for each dice and both dices are thrown together. The total number of outcomes for each of the cases is 36. For finding the number of favourable outcomes check for each and every case manually. For example: the number of cases of doublet are (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6), i.e., six favourable cases.

Complete step by step solution:
Before moving to the question, let us talk about probability.
Probability in simple words is the possibility of an event to occur.
Probability can be mathematically defined as $=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$ .
Start by drawing the table of all the possible outcomes. There would be a total of 36 outcomes, as 6 options for each dice and both dices are thrown together. The total number of outcomes for each of the cases is 36.

Let us start with sub part (a). The favourable outcomes, i.e, the sum as 8 are (4,4), (6,2), (2,6), (3,5), (5,3). So, the number of favourable outcomes are 5.
$\text{P}\left( \text{sum equals 8} \right)=\dfrac{5}{36}$
Now, let us move to subpart (b). The favourable outcomes of a doublet are: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). So, 6 favourable outcomes.
$\text{P}\left( \text{a doublet} \right)=\dfrac{6}{36}=\dfrac{1}{6}$
In subpart (c), we are asked about doublet primes. So, the primes on a dice are 2, 3, 5. So the favourable outcomes are: (2,2), (3,3), (5,5), i.e., 3 favourable outcomes.
$\text{P}\left( \text{a doublet of prime} \right)=\dfrac{3}{36}=\dfrac{1}{12}$
In subpart (d), we are asked about doublet of odd numbers. So, the odd numbers on a dice are 1, 3, 5. So the favourable outcomes are: (1,1), (3,3), (5,5), i.e., three favourable outcomes.
$\text{P}\left( \text{a doublet of odd numbers} \right)=\dfrac{3}{36}=\dfrac{1}{12}$
Now the subpart (e) is about those cases where the sum of the numbers on the two dice is greater than 9. The favourable outcomes for this case are: (6,5), (6,4), (5,5), (5,6), (4,6), (6,6).
$\text{P}\left( \text{sum greater than 9} \right)=\dfrac{6}{36}=\dfrac{1}{6}$

Note: Whenever you are asked the probabilities related to 2 dices, it is better to draw the table of outcomes, because it gives you the clear idea of the situation and doesn’t consume more time also. However, when three or more dice are used, it is not possible to draw the table because there would be 216 or more outcomes to it. Also, make sure that you don’t miss any outcomes while noting the favourable outcomes manually.