Two dice are holled. Find the of probability that the sum of outcomes is (i) equal to 4 (ii) greater than 10 (iii) less than 13
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Answers
Assumptions
The dice are “ fair “, that is, not biassed in any manner.
The dice are both six-sided dice, that is both have 6 faces, with each face on each dice, showing one of the numbers, 1 to 6, with no number repeated on the same dice.
Analysis
With two dice, there are ( 6 ) * ( 6 ) = ( 36 ) possible combinations of numbers.
The minimum sum possible for the two dice thrown is (1, 1) = a sum of (2 )
The maximum sum possible for the two dice thrown is (6, 6) = a sum of (12).
Sum = (1).
The minimum possible sum is (1, 1) = ( 2 ).
Therefore P( 1 ) = ( 0 )/( 36 ) = 0
Sum = (4)
A sum of (4) can be achieved with number combinations ( 2, 2), (1, 3) and (3, 1), that is only with 3 combinations of numbers.
P(sum = 4) = ( 4 / 36 ) = ( 1 / 9 )
Sum = (less than 13 )
To meet this requirement:
0 < sum =/< 12
The combinations which meet this requirement are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6 )
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5) (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).
which gives a total of 36 combinations.
Note however we needn't have written out all the possible combinations as I did above. All I needed to do, to calculate the.probability that the sum was less than ( 13 ) was to recognize that a sim of (1) to (12) inclusive included every possible combinations of numbers from two dice.
P(sum < 13) = ( 36 / 36 ) = 1 = 100%
PB
Step-by-step explanation:
n(s) = 6*6 = 36
(1) E1 = equals to 4
n(E1) = {(2,2),(1,3),(3,1)} = 3
P(E1) = n(E1)/n(s)
= 3/36
= 1/12
(2)E2 = greater than 10
n(E2) = {(6,6),(5,6),(6,5)} = 3
P(E2) = 1/12
(3)E3 = less than 13
since the maximum no. which can sum up is 12 so n(E3) = 36
P(E3) = 1