Conditional probablility P(A/B) = P(A and B) / P(B). Here, A is sum of two dice being greater than or equal to 9 and B is at least one of the dice showing 6. Number of ways two dice faces can sum up to 9 = (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 10 ways. Number of ways that at least one of the dice must show 6 = (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1) = 11 ways. Number of ways of rolling a number greater than or equal to 9 and at least one of the dice showing 6 = (3, 6), (4, 6), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 7 ways. Probability of rolling a number greater than or equal to 9 given that at least one of the dice must show a 6 = 7 / 11
Answer:
Step-by-step explanation:
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f(3) is equal to -101.
<u>Step-by-step explanation</u>:
- The sequence is followed by f(0)=4
- The given expression is f(n+1)= -3f(n)+1
Put n=0,
f(0+1)= -3f(0)+1
f(1)= -3(4)+1
f(1)= -12+1 = -11
Put n=1,
f(1+1)= -3f(1)+1
f(2)= -3(-11)+1
f(2)= 33+1 = 34
Put n=2,
f(2+1)= -3f(2)+1
f(3)= -3(34)+1
f(3)= -102+1
f(3)= -101
