EE 422: Digital Communications Home work # 1 1) A player has a loaded die which

Question

EE 422: Digital Communications Home work # 1 1) A player has a loaded die which turns up the number I wit 2 to 6 with probability 1/15 each. Unfortunately, he left his loaded honest dice and couldn't tell them apart. He picked one die (at random) from the box h probability 2/3 and the number die in a box with two ed it once, and the number 1 appeared. Conditional on this res he picked up the honest die? The player rolled the die once more, and it came up ult, what is the probability that 1 again. hat is the probability after this second rolling that he has picked up the honest die?

Explanation / Answer

Probability of getting a 1 from the loaded die is given as:

P( 1 | loaded ) = 2/3

P( 1 | honest ) = 1/6 as all numbers in this are equally likely.

Also as there are 2 honest dies and 1 loaded dice in the box, we get:

P( loaded ) = 1/3 and P( honest ) = 2/3

Using the law of total probability, we get the total probability of getting a 1 as:

P( 1) = P( 1 | honest )P( honest ) + P( 1 | loaded )P( loaded )

P( 1) = (1/6)*(2/3) + (2/3)*(1/3)

P( 1) = (1/9) + (2/9) = (1/3)

Given that a 1 came, probability that he picked up an honest die is computed as: ( Using Bayes theorem ) :

P( honest | 1 ) = P( 1 | honest )P( honest ) / P(1) = (1/6)*(2/3) / (1/3) = 1/3

Therefore (1/3) = 0.3333 is the required probability here.

Now for the second part of the problem:

Probability of getting a double 1 from the 2 types of dies is given as:

P( double 1 | loaded ) = (2/3)*(2/3) = (4/9)

P( double 1 | fair ) = (1/6)*(1/6) = (1/36)

Therefore using law of total probability we get:

P( double 1 ) = P( double 1 | loaded ) P( loaded ) + P( double 1 | fair ) P(fair )

P( double 1 ) = (4/9)*(1/3) + (1/36)*(2/3) = (4/27) + (1/54) = (1/6)

Using the Bayes theorem, given that we got double 1, probability that we picked up the honest die is computed as:

P( fair | double 1) = P( double 1 | fair ) P(fair ) / P( double 1) = (1/36)*(2/3) / (1/6) = (1/9 )

Therefore the required probability here is (1/9) = 0.1111

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