Question

In: Statistics and Probability

Real numbers p and q are randomly chosen from the interval 0 to 1, inclusive. If...

Real numbers p and q are randomly chosen from the interval 0 to 1, inclusive. If r is given by r = 2(p + q), and p, q, r are rounded to the nearest integers to give P, Q and R, respectively, determine the probability that R = 2(P + Q). (As an example, if p=0.5 and q=0.381, then r=1.762, and so P =1, Q=0, and R=2)

Solutions

Expert Solution

Answer :

R = 2(P+Q) is true only if 2(P+Q) is equal to rounded off value of r. So we will look into 4 cases. In each case we will find the value of R if R = 2(P+Q) and probability that rounded of value of r is equal to THAT R.

Case 1:

which implies P = 0

which implies Q = 0

which implies but R = 2(0+0) = 0

r = R = 0 when . Because r is evenly distributed since both p and q are evenly distributed the probability of r falling in this region out of all the possible values it can take upto 2 is

Case 2:

which implies P = 0.

which implies Q = 1.

which implies but R = 2(0+1) = 2

r = R = 2 when . Because r is evenly distributed since both p and q are evenly distributed the probability of r falling in this region out of all the possible values it can between 1 and 3 is

Case 3:

which implies P = 1.

which implies Q = 0.

which implies but R = 2(1+0) = 2

r = R = 2 when . Because r is evenly distributed since both p and q are evenly distributed the probability of r falling in this region out of all the possible values it can between 1 and 3 is

Case 4:

which implies P = 1.

which implies Q = 1.

which implies but R = 2(1+1) = 4

r = R = 4 when . Because r is evenly distributed since both p and q are evenly distributed the probability of r falling in this region out of all the possible values it can between 2 and 4 is .

Thus combining the probabilities from all the 4 cases and multiplying by the probability of that particular case happening, we get

Probability of R = 2(P+Q) is

= where a is the probability of R = 2(P+Q) in case i and b is the probability of the occurence of the case i which is equal i.e. 1/4 for all four cases.

= 1/4 X 1/4 + 1/2 X 1/4 + 1/2 X 1/4 + 1/4 X 1/4

= 1/16 + 1/8 + 1/8 + 1/16

= 6/16 = 3/8

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