Q. A black die and a white die are rolled. Find the probability that the number shown by the black die will be more than twice that shown by the white die.
BITSATBITSAT 2010
Solution:
The number of favourable cases are shown below:
1
3
1
4
1
5
1
6
2
5
2
6
There are $6$ favourable cases in which the number on black die is more than twice the number on the white die.
$\therefore m = 6$
$n =$ Total number of cases $= 6 \times 6$
( $\therefore $ with each die there are six possibilities)
$\therefore $ Probability $p = \frac{m}{n} = \frac{6}{ 6 \times 6 } = \frac{1}{6} $
| 1 | 3 |
| 1 | 4 |
| 1 | 5 |
| 1 | 6 |
| 2 | 5 |
| 2 | 6 |