The newest invention of the 6.041x staff is a three-sided die. On any roll of this die, the result is 1 with probability 1/2, 2 with probability 1/4, and 3 with probability 1/4.

Consider a sequence of six independent rolls of this die.

1. Find the probability that exactly two of the rolls results in a 3.

2. Given that exactly two of the six rolls resulted in a 1, find the probability that the first roll resulted in a 1.

3. We are told that exactly three of the rolls resulted in a 1 and exactly three rolls resulted in a 2. Given this information, find the probability that the six rolls resulted in the sequence (1,2,1,2,1,2).

4. The conditional probability that exactly k rolls resulted in a 3, given that at least one roll resulted in a 3, is of the form:
11−(c1/c2)c3(c3k)(1c2)k(c1c2)c3−k,for k=1,2,…,6.


Find the values of the constants c1, c2, and c3:

part-2

ans:= 1/3

1. (6 2)*(1/4)^2*(3/4)^4

3. 0.05

4. c1=3, c2= 4, c3=6

1 Given that exactly two of the six rolls resulted in a 1, find the probability that the first roll resulted in a 1. so is the solution 1/3

2 We are told that exactly three of the rolls resulted in a 1 and exactly three rolls resulted in a 2. Given this information, find the probability that the six rolls resulted in the sequence (1,2,1,2,1,2) . is 1/20

3 he conditional probability that exactly k rolls resulted in a 3, given that at least one roll resulted in a 3, is of the form:

11−(c1/c2)c3(c3k)(1c2)k(c1c2)c3−k,for k=1,2,…,6.

Find the values of the constants c1 =3 , c2=4 , and c3 =6:

1. To find the probability that exactly two of the rolls result in a 3, we can use the binomial probability formula. The probability of getting exactly k successes (in this case, rolling a 3) in n trials (in this case, 6 rolls) is given by:

P(X = k) = (nCk) * (p^k) * (q^(n-k))

where nCk represents the binomial coefficient, p is the probability of success (rolling a 3), q is the probability of failure (rolling a 1 or a 2), and k ranges from 0 to n.

So, the probability of exactly two rolls resulting in a 3 is:

P(X = 2) = (6C2) * (1/4)^2 * (3/4)^(6-2)

We can calculate this value to get the actual probability.

2. Given that exactly two of the six rolls resulted in a 1, we can treat those two rolls as fixed and calculate the probability of the first roll resulting in a 1. This probability is simply the conditional probability of the first roll being a 1, given that exactly two rolls resulted in a 1. This conditional probability will be equal to the total probability of getting two 1s divided by the total probability of getting exactly two 1s.

So, to find this probability, we first need to calculate the total probability of getting exactly two 1s.

P(exactly two 1s) = (6C2) * (1/2)^2 * (1/2)^(6-2)

Then, we divide this value by the total probability of getting two 1s:

P(first roll is a 1 | exactly two 1s) = P(exactly two 1s) / P(two 1s)

By calculating these probabilities, we can find the desired conditional probability.

3. Given that exactly three of the rolls resulted in a 1 and exactly three rolls resulted in a 2, the probability of the six rolls resulting in the sequence (1,2,1,2,1,2) is simply the product of the individual probabilities for each roll.

P(1,2,1,2,1,2) = (1/2) * (1/4) * (1/2) * (1/4) * (1/2) * (1/4)

We can compute this value to find the probability.

4. To find the values of the constants c1, c2, and c3 in the given conditional probability formula, we need to examine the information and understand the meaning of these constants.

In the formula 11−(c1/c2)c3(c3k)(1c2)k(c1c2)c3−k, for k=1,2,…,6, "k" represents the number of rolls resulting in a 3, and "c1," "c2," and "c3" are various constants.

Without further information, it is not possible to determine the values of c1, c2, and c3. These constants seem to represent some probabilities or coefficients that may depend on the specific problem or context.

1. To find the probability that exactly two of the six rolls result in a 3, we can use the binomial distribution. The binomial distribution calculates the probability of getting a certain number of successes in a fixed number of trials.

In this case, the probability of getting a 3 on any roll is 1/4. So, the probability of not getting a 3 on any roll is 1 - 1/4 = 3/4.

We want exactly two rolls to result in a 3, which means we have 4 rolls that don't result in a 3. The probability of getting exactly two 3's and four non-3's can be calculated as:

(6 choose 2) * (1/4)^2 * (3/4)^4

where (6 choose 2) represents the number of ways you can choose 2 rolls out of 6.

2. Given that exactly two of the six rolls resulted in a 1, we want to find the probability that the first roll resulted in a 1.

Since we know that exactly two rolls resulted in a 1, there are a total of 5 rolls remaining. Out of these 5 rolls, we want to find the probability that the first roll is a 1. So, the probability can be calculated as:

1/5

This is because each roll is independent, and the probability of rolling a 1 on any roll is 1/2.

3. We are told that exactly three of the rolls resulted in a 1 and exactly three rolls resulted in a 2. Given this information, we want to find the probability that the six rolls resulted in the sequence (1,2,1,2,1,2).

Since each roll is independent, the probability of obtaining this specific sequence can be calculated by multiplying the individual probabilities of each roll.

The probability of rolling a 1 is 1/2, and the probability of rolling a 2 is 1/4. So, the probability of getting the sequence (1,2,1,2,1,2) can be calculated as:

(1/2)^3 * (1/4)^3

4. The given conditional probability is in the form:
11−(c1/c2)c3(c3k)(1c2)k(c1c2)c3−k, for k=1,2,…,6.

To find the values of the constants c1, c2, and c3, we need to match this form with the given conditional probability.

Comparing the form with the given conditional probability, we can derive the following relationships:

c1 = number of rolls resulting in a 3
c2 = total number of rolls (6 in this case)
c3 = number of rolls not resulting in a 3 (c2 - c1)

By substituting these values, we can solve for the constants c1, c2, and c3.