Test Index

BITSAT 2020 Paper

Section: Mathematics

Question : 120 of 150

Marks: +1, -0

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

Validate

Solution:

Total number of ways

=6 \times 6 \times \dots .

n

times

=6^{n}

Total number of ways to show only odd number

=3 \times 3 \times \dots

n

times

=3^{n}

\therefore

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}

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