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

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) 6n-1 (B) 3n-1 (C) 6n-3n (D) None of these

Tardigrade Admin · Accepted Answer

Total number of ways =6 × 6 × ldots to n times =6n Total number of ways to show only even numbers =3 × 3 × ldots n times -3n Therefore, the required number of ways =6n-3n

Q. 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

$6^{n} - 1$

$3^{n} - 1$

$6^{n} - 3^{n}$

None of these

Total number of ways $= 6 \times 6 \times \dots$ to $n$ times $= 6^{n}$
Total number of ways to show only even numbers $= 3 \times 3 \times \dots n$ times $- 3^{n}$
Therefore, the required number of ways $= 6^{n} - 3^{n}$

Q. 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

6n−1

3n−1

6n−3n