Roll a 4-sided die twice and assume all sixteen outcomes are equally likely. Con

ID: 2989714 • Letter: R

Question

Roll a 4-sided die twice and assume all sixteen outcomes are equally likely. Consider the following events:

A : The difference of the two numbers is 2 (Let x denote the first number and y denote the second number. We have x ? y = 2 or y ? x = 2.)

B:At least one of the numbers is2.

C : The two numbers are equal.

Compute the following probabilities P (A ? B), P (A ? C), P (C|B) and P (A|B).

A conservative design team C and an innovative design team I are asked to separately design a

new product within a month. From the past experience, we know

The probability that team I is successful is 1/2

The probability that one and exactly one team is successful is 5/8

The probability that both teams are successful is 1/4

Conditioned on at least one team is successful, what is the probability that team C fails?

events are as follows:

A: the difference of 2 numbers is 2.
{(1,3),(3,1),(2,4),(4,2)}

B: at least one of the number is 2.
{(2,1),(2,2),(2,3),(2,4),(1,2),(3,2),(4,2)}

C: two numbers are equal.

{(1,1),(2,2),(3,3),(4,4)}

P(AUB)=9/16

P(A intersection C)=0

P(C/B)=P(C intersection B)/P(B)=(1/16)/(7/16)=1/7

2. let A: team C is successful
B: team I is successful.

so P(B)=1/2

P( [A intersection B' ] U [ A' intersection B])=5/8

==> P(AUB)-P(A intersection B)=5/8

P(A intersection B)=1/4

so P(AUB)=7/8

we have to find

P(A'/(AUB))=P(A' intersection (AUB))* P(AUB)

=P(B)*P(AUB)=1/2 * (7/8)=7/16

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