Question

(3) Every day, during his lunch break, an undergraduate student of probability repeatedly rolls a balanced die until an odd prime number turns up for the first time. If X = the number of rolls he needs to get the first odd prime number, what is E(X)? Suppose that today he has already rolled it 6 times and hasn’t gotten an odd prime number yet. What are the conditional chances that he will need at least 10 attempts (including the 6 he has already made) to be successful?

Answer 1

since odd prime numbers are 3,5

therefore probability of getting an odd prime number on a die =p=2/6 =1/3

the number of rolls till a odd prime appears follow a geometric distribution with parameter p=1/3

therefore E(X) =1/p=1/(1/3) =3 rolls

2)

P(no odd prime in first 6 rolls )= P(X>6)=(1-1/3)⁶ =(2/3)⁶

P(at least 10 attempts required) =P(no odd prime in first 9 rolls )=P(X>9)=(1-1/3)⁹ =(2/3)⁹

therefore P(at least 10 rolls are required given no odd prime in first 6)

=P(X>=10|X>6) =(2/3)⁹/(2/3)⁶ =(2/3)³ =8/27