The newest invention 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. c1=3, c2= 4, c3=6
1. (6)(1/4)^2 *(3/4)^4
(2)
2. 1/3
3. 1/20
1. To find the probability that exactly two of the rolls result in a 3, we can use the binomial probability formula. The formula is given by:
P(X = k) = (nCk) * p^k * (1-p)^(n-k)
Where:
n = number of trials (6 rolls)
k = number of successful trials (2 rolls resulting in a 3)
p = probability of success on a single trial (probability of rolling a 3, which is 1/4)
Plugging in the values:
P(X = 2) = (6C2) * (1/4)^2 * (3/4)^(6-2)
P(X = 2) = (15) * (1/16) * (9/16)
P(X = 2) = 135/256
Therefore, the probability that exactly two of the rolls result in a 3 is 135/256.
2. Given that exactly two of the six rolls resulted in a 1, we need to find the probability that the first roll resulted in a 1, assuming that condition. Since there are six rolls and exactly two need to result in a 1, we will use the binomial distribution again.
P(First roll = 1 | X = 2) = P(First roll = 1 & X = 2) / P(X = 2)
P(First roll = 1 & X = 2) = (5C1) * (1/2)^1 * (1/4)^2 * (1/4)^3
P(First roll = 1 & X = 2) = (5) * (1/2) * (1/16) * (1/64)
P(First roll = 1 & X = 2) = 5/2048
P(X = 2) = 135/256 (as calculated in question 1)
P(First roll = 1 | X = 2) = (5/2048) / (135/256)
P(First roll = 1 | X = 2) ≈ (5/2048) * (256/135)
P(First roll = 1 | X = 2) ≈ 5/27
Therefore, the probability that the first roll resulted in a 1, given that exactly two of the six rolls resulted in a 1, is approximately 5/27.
3. Given that exactly three rolls resulted in a 1 and exactly three rolls resulted in a 2, we need 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 each specific sequence is given by the product of the probabilities of each individual roll.
P(1,2,1,2,1,2) = (1/2) * (1/4) * (1/2) * (1/4) * (1/2) * (1/4)
P(1,2,1,2,1,2) = (1/2)^3 * (1/4)^3
P(1,2,1,2,1,2) = 1/64
Since there are multiple sequences that can result in three 1s and three 2s, we need to consider all possible permutations. The number of permutations is given by the multinomial coefficient:
(6C3,3) = (6!)/(3!*3!)
(6C3,3) = (6*5*4)/(3*2)
(6C3,3) = 20
Therefore, the probability that the six rolls resulted in the sequence (1,2,1,2,1,2), given that exactly three rolls resulted in a 1 and exactly three rolls resulted in a 2, is 1/64 * 20 = 20/64 = 5/16.
To answer these probability questions, we can use the concept of probability distribution and apply the rules of probability. Let's break down each question step by step:
1. Find the probability that exactly two of the rolls result in a 3:
To find the probability of this event, we need to calculate the probability of getting exactly two 3's in six rolls. Since each roll is independent, we can use the binomial distribution to solve this problem.
The probability of getting exactly two 3's can be calculated using the formula:
P(X=k) = (nCk) * (p^k) * ((1-p)^(n-k))
Here, n is the number of trials (6 rolls), k is the number of successes (2 rolls resulting in a 3), and p is the probability of success (1/4 for rolling a 3).
Using the formula, we get:
P(X=2) = (6C2) * ((1/4)^2) * ((1-(1/4))^(6-2))
= (15) * (1/16) * (3/4)^4
= 15 * (1/16) * (81/256)
= 1215/4096
Therefore, the probability that exactly two of the rolls result in a 3 is 1215/4096.
2. Given that exactly two of the six rolls resulted in a 1, find the probability that the first roll resulted in a 1:
Since we know exactly two rolls resulted in a 1, we need to determine the probability that the first roll was a 1.
Let's calculate this using a conditional probability. The conditional probability is given by the formula:
P(A|B) = P(A and B) / P(B)
Here, A represents the event of the first roll resulting in a 1, and B represents the event of exactly two rolls resulting in a 1.
We need to find P(A and B) and P(B).
P(A and B) is the probability that the first roll is 1 and exactly two rolls are 1, which is the same as the probability of getting a 1 on the first roll times the probability of getting exactly one 1 in the next five rolls.
P(A and B) = (1/2) * (5C1) * ((1/2)^1) * ((1/2)^4)
= 5/32
P(B) is the probability of getting exactly two 1's in six rolls. We can calculate this using the binomial distribution formula as before:
P(B) = (6C2) * ((1/2)^2) * ((1-(1/2))^(6-2))
= (15) * (1/4) * (1/2)^4
= 15/64
Now, we can substitute these values into the formula for conditional probability:
P(A|B) = P(A and B) / P(B)
= (5/32) / (15/64)
= 5/8
Therefore, the probability that the first roll resulted in a 1, given that exactly two of the six rolls resulted in a 1, is 5/8.
3. Given that exactly three rolls resulted in a 1 and exactly three rolls resulted in a 2, find the probability that the six rolls resulted in the sequence (1,2,1,2,1,2):
To find the probability of a specific sequence occurring, we can multiply the independent probabilities of each individual roll.
Given that we have exactly three 1's and exactly three 2's, the probability of any specific six-roll sequence satisfying this condition is:
P = (1/2) * (1/4) * (1/2) * (1/4) * (1/2) * (1/4)
= 1/1024
Therefore, the probability that the six rolls resulted in the sequence (1,2,1,2,1,2), given that exactly three rolls resulted in a 1 and exactly three rolls resulted in a 2, is 1/1024.