Four fair dice, colored red, green, blue, and white, are tossed. (a) Determine t

Question

Four fair dice, colored red, green, blue, and white, are tossed.

(a) Determine the probability of getting all four face values equal to 3.

(b) After tossing, a quick glance at the outcome indicated that two of the face values were 3 but no other information (about their color or the values of the remaining two faces) was noted. Now determine the probability of getting all four face values equal to 3.

(c) After tossing it was further noted that, out of the two observed face values of 3, one was red in color. Now determine the probability of getting all four face values equal to 3.

(d) After tossing, it was finally confirmed that the two observed face values of 3 were red and white. Now determine the probability of getting all four face values equal to 3.

(e) You should get distinct answers for the above four probabilities. Qualitatively explain why the above four probabilities make sense.

These are supposed to be the correct answers: (a) 1/1296 (b) 1/171, (c) 1/91, and (d) 1/36.

Explanation / Answer

When we rolled 4 dices possible number of outcomes are 6*6*6*6 = 1296

(a)

Out of these 1296 outcomes, 1 outcome has all three so

P(all three) = 1/1296

(b)

Number of outcomes with two 3's:

Number of ways of selecting 2 places for 2 3's is C(4,2) = 6

Rest two position can be filled in 5*5 = 36 ways

Number of outcomes with three 3's:

Number of ways of selecting 3 places for 3 3's is C(4,3) = 4

Rest two position can be filled in 5 = 5 ways

That is number of ways of outcome with 3 3's is: 5 * 4= 20

Number of outcomes with four 3's:

That is number of ways of outcome with 4 3's is: 1

So number of outcomes with at least 2 3's is: 150+20+1=171

Out of these 171 outcomes, only one has all three so the probability of getting all four face values equal to 3 given that two of the face values were 3 is

P(2 face values are 3 | all 3) = 1/ 171  

(c)

Red die has 3. Now other three can be one on other three dices.

Number of outcomes with one 3' in rest three dies:

C(3,1) * 5*5 =75

Number of outcomes with two 3' in rest three dies:

C(3,2) * 5 =15

Number of outcomes with three 3' in rest three dies:1

So number of outcomes with at least 2 3's such that one 3 in on red die is: 15+75 +1 =91

That is the probability of getting all four face values equal to 3 given that out of the two observed face values of 3, one was red in color is

P(all three | 2 3's with one 3 on red) = 1/91

(d)

Red and white die has 3s. Rest two die can shows 6*6 = 36 values. So possible number of outcomes with red and white dice having 3 is: 36

That is the probability of getting all four face values equal to 3 given that out of the two observed face values of 3, one was red in color and other is white is

P(all three | 2 3's with one 3 on red and other on white) = 1/36

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