Suppose that a die is loaded so that {1, 3, 4, 6} each come up equally often as
ID: 3819241 • Letter: S
Question
Suppose that a die is loaded so that {1, 3, 4, 6} each come up equally often as each other, 2 comes up three times as often {1, 3, 4, 6} and 5 comes up twice as often as {1, 3, 4, 6}. Answer the following questions about this die. Show all work. (a) Find the probability distribution p for this loaded die. (b) What is the probability that you roll an odd number with the die? (c) What is the probability that you roll the die twice and the two rolls sum to 7? (d) What is the probability that you roll the die twice and the two rolls sum to 7 given that the first roll is an even number? (e) Based on your answers to parts (c) and (d), are the events that you roll an even on the first roll and the two dice sum to 7 independent?
Explanation / Answer
Let p be probability of getting 1 or 3 or 4 or 6. Probability of getting 2 is 3p, and that of getting 5 is 2p.
i.e. p + p + p + p + 3p + 2p = 1 ( summation has to be one )
p = 1/9
a) Probability Distribution of Die =>
Probability of getting { 1,3,4,6 } = 1/9
Probability of getting 2 = 3/9
Probability of getting 5 = 2/9
b) P( Odd Number ) = P( X = 1 ) + P( X = 3 ) + P( X = 5 ) = 1/9 + 1/9 + 2/9 = 4/9 ( we can do summation as e7nts are independent )
c) P( Sum of two rolls is 7 ) = P( X=1,Y=6 ) + P( X=6,Y=1 ) + P(X=2,Y=5) + P(X=5,Y=2) + P(X=3,Y=4) + P(X=4,Y=3)
= 1/81 + 1/81 + 6/81 + 6/81 + 1/81 + 1/81 = 16/81
d) Probability is = P( X=6,Y=1 ) + P(X=2,Y=5) + P(X=4,Y=3)
= 8/81 ( half of previous )
e) If they are independent, then P ( Sum of two rolls is 7, and first roll is even ) should be equal to
P( Sum of rolls is 7 )*P( first roll is even )
Note that P( Roll is even ) = 1 - P( Roll is odd ) = 1 - 4/9 ( from (b) ) = 5/9
i.e. 8/81 = (16/81) *5/9 CONTRADICTION
So, now these events are not independent
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