How can I calculate the probability of X = 1,2,3 in a three-toss die experiment?

[leptons]

Consider the experiment of tossing a die thrice. X is defined as the number of different faces that appear (i.e., X = 1,2,3). What is meant by the "number of different faces that appear"? Could you help me how could I get P(X = 1,2,3)?

 

[TenaliRaman]

well,
u r rolling three dice ... so u could get them all different so number of different faces u will have is 3 ... or you could get 2 same numbers but 1 different , so number of different faces u see is 2 ... or u could get all the three numbers same which means number of different faces u see is 1.

so X can take values 1,2 and 3.

Now u can solve it pretty easily in two possible ways ,
1> enumerate all possible triplets
(1,1,1),(1,1,2),...(6,6,6).
then simply do the counting.

2> if you are good at combinatorics ... then u can do it faster by considering
a. number of ways u can get all three numbers different
b. number of ways u can get two numbers same and 1 different
c. number of ways u can get all three numbers same

-- AI

 

[tacman]

The "number of different faces that appear" refers to the number of unique outcomes that occur when the die is tossed three times. In other words, if the die lands on the same face all three times, the number of different faces that appear would be 1. If it lands on two different faces, the number of different faces that appear would be 2, and so on.

To calculate P(X = 1,2,3), we need to first determine the total number of possible outcomes. In this case, there are 6 possible outcomes for each toss, giving us a total of 6*6*6 = 216 possible outcomes for three tosses.

Next, we need to determine the number of outcomes that result in X = 1,2,3. This can be done by listing out all the possible combinations of three tosses that would result in 1, 2, or 3 different faces appearing. For example, if the first toss results in a 1, the second in a 2, and the third in a 3, then X = 3. Similarly, if the first toss results in a 1, the second in a 1, and the third in a 2, then X = 2.

After listing out all possible combinations, we can see that there are 56 outcomes that result in X = 1, 84 outcomes that result in X = 2, and 36 outcomes that result in X = 3. Therefore, P(X = 1,2,3) = (56+84+36)/216 = 0.463 or approximately 46.3%.