TWo dice are numbered 1,2,4,4,5,6 and 1,1,2,2,3,3 respectively.They are thrown and some of the numbers on them are noted.Find the probability of getting each sum from 2 to 9 separately

Question

TWo dice are numbered 1,2,4,4,5,6 and 1,1,2,2,3,3 respectively
They are thrown and some of the numbers on them are noted Find the
probability of getting each sum from 2 to 9 separately

Manish Bharti · Accepted Answer

Total number of possible outcomes $= 6 \times 6 = 36$
1. The outcomes favourable to the the event 'the sum of the two numbers is 2' denoted by $E_{1}$ , are (1,1) and (1,1)
Since 2nd dice is numbered 1 on two sides, therefore both 1's will be considered as a different outcome.
$P (E_{1}) = \frac{2}{36} = \frac{1}{18}$

Other way to understand this is -
Probability of getting 1 in 1st dice = 1/6
Probability of getting 1 in 2nd dice = 2/6
$P (E_{1}) = \frac{1}{6} \times \frac{2}{6} = \frac{1}{18}$

2. The outcomes favourable to the the event 'the sum of the two numbers is 3' denoted by $E_{2}$ , are (1,2), (1,2), (2,1) and (2,1)
$P (E_{2}) = \frac{4}{36} = \frac{1}{9}$

3. The outcomes favourable to the the event 'the sum of the two numbers is 7' denoted by $E_{3}$ , are (4,3),(4,3),(4,3),(4,3), (5,2), (5,2), (6,1) and (6,1)
$P (E_{2}) = \frac{8}{36} = \frac{2}{9}$

Similarly you can solve other parts