What is the probability of getting consecutive 6,6 before consecutive 6,5?
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The probability of getting consecutive 6,6 is zero (0)
Step-by-step explanation:
When two dice are thrown simutaneously.
Total favourabe outcomes 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).
Total number of favourabe outcomes = 36
Nuber of favourable outcome = 0
P(E) = 0/36 =0
Answered by
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Answer:
0
Step-by-step explanation:
The event of getting 6,6 and event of getting 6,5 are mutually independent of each other. This means occurring of one does not depend upon the other.
So, the probability of getting consecutive 6,6 before consecutive 6,5 is zero.
It is actually 0/36 = 0
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