Question

The number of possible outcomes in a throw of n ordinary dice in which at least one of the dice shows an odd number is
(a) $$6^n-1$$
(b) $$3^n-1$$
(c) $$6^n-3^n$$
(d) None of these

Answer 1

Hint: First of all, find the total outcomes when ‘n’ dice are thrown. Now, find the number of outcomes when there would be no odd number or all even numbers. Now subtract this from the total outcomes to get the desired answer.

Complete step-by-step answer:
In this question, we have to find the number of possible outcomes in a throw of n ordinary dice in which at least one of the dice shows an odd number. Let us see what dice is and what outcomes are.
Dice: Dice is a small cube with each side having a different number of spots on it ranging from one to six that is 1, 2, 3, 4, 5, and 6 dots. We generally use this in the games involving chance. The dice is rolled or thrown and dice comes to rest showing a random number from 1 to 6. Also, as each has different dots/numbers. So, each one is equally likely to come when the die is rolled.
Outcomes: An outcome is a result of an event that depends on probability and any event can have more than one possible outcome. For example, when we toss a coin, we can get either the head or tail. So there are a total of 2 outcomes.
Now, let us consider our question. We know that if we throw a die, any of the 6 faces having 1, 2, 3, 4, 5 or 6 dots can come. So, there are a total of 6 outcomes when we throw a die. Again, when we throw a second die, there would be 6 outcomes. Similarly, when we throw any of the die out of n dice individually, there would be 6 outcomes. So, we get the total possible outcomes in a throw of n dice simultaneously
$$=6\times 6\times 6.....n\text{ times}$$
$$={\left(6\right)}^n$$
Now, let us find the total number of outcomes when n dice show odd numbers or in other words when all the dice show even numbers. When we throw a die, out of six numbers, 3 numbers that are 2, 4, and 6 are even. So, there would be 3 outcomes for each die.
So, when n dice are thrown simultaneously, then the total number of possible outcomes when no dice shows the odd number or all dice show even number
$$=3\times 3\times 3.....n\text{ times}$$
$$={\left(3\right)}^n$$
Now, we have to find the outcomes in which at least one of the dice shows an odd number. In other words, we have to subtract the possible outcomes when dice shows odd numbers from the total possible outcomes. So, we get,
The total number of possible outcomes in which at least one dice shows an odd number = (Total possible outcomes – possible outcomes when dice shows an odd number)
$$=6^n-3^n$$
Hence, the option (c) is the right answer.

Note: In the questions of permutation and combinations or probability, students must note that whenever we are asked outcomes of a particular thing, we do it in a reverse way and instead of finding it directly, we subtract outcomes of none of that particular from total outcomes. For example, in the above question, we were asked to find the outcomes when at least one of the dice shows an odd number, so we subtracted outcomes when no dice showed odd numbers from total outcomes.