SOLUTION: Suppose that you are performing the probability experiment of rolling one fair six-sided die. Let F be the event of rolling a four or a five. You are interested in how many times y

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Question 1206756: Suppose that you are performing the probability experiment of rolling one fair six-sided die. Let F be the event of rolling a four or a five. You are interested in how many times you need to roll the die in order to obtain the first four or five as the outcome.
• p = probability of success (event F occurs)
• q = probability of failure (event F does not occur)
Find the values of p and q. (Enter exact numbers as integers, fractions, or decimals.)
Find the probability that the first occurrence of event F (rolling a four or five) is on the second trial. (Round your answer to four decimal places.)

Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.
Suppose that you are performing the probability experiment of rolling one fair six-sided die.
Let F be the event of rolling a four or a five. You are interested in how many times you need
to roll the die in order to obtain the first four or five as the outcome.
• p = probability of success (event F occurs)
• q = probability of failure (event F does not occur)
(a) Find the values of p and q. (Enter exact numbers as integers, fractions, or decimals.)
(b) Find the probability that the first occurrence of event F (rolling a four or five)
is on the second trial. (Round your answer to four decimal places.)
~~~~~~~~~~~~~~~~~~~~~~~~~

(a)  p = probability that " a four or a five" occurs (event F).  This probability is

         p =  +  =  = .    


     q = probability that event F does not occur.  It is a complementary event to F.

         q = 1 - p = 1 -  = .



(b)  Probability of (b) is 

         P = P(event F is not a first roll; event F is the second roll) = (1-p)*p = q*p =  = .

Solved.



Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


P(4 or 5) = 4/6 = 2/3

P(other) = 1-2/3 = 1/3

P(first 4 or 5 on 2nd throw) = P((other on 1st throw) AND (4 or 5 on 2nd throw)) = (2/3)*(1/3) = 2/9


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