The probability of seeing at least 6 dots, but less than 10 dots when 2 fair dice are rolled is 20/36 or 5/9.
Probability is the ratio of favorable outcomes to the total possible outcomes of an event.
P(X) =n(X)/n(S)
Where the total possible outcomes are given by the sample space S and the event is X.
When two dice are rolled, the possible outcomes are 36.
The sample space is as follows:
{(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)}
I.e., n(S) = 36
From the about the favorable outcomes of the event - the dice showing at least 6 but less than 10 dots (sum of dots on two dice) are:
{(1,5), (1,6), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6), (4,2), (4,3), (4,4), (4,5), (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3)}
Thus, there are 20 such favorable outcomes. I.e., n(X) = 20
Then, the required probability is
P(X) = n(X)/n(S)
= 20/36
= 5/9
Therefore, the required probability is 5/9.
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