You roll a fair six-sided die (all 6 of the possible results of a die roll are equally likely) 5 times, independently. Let X be the number of times that the roll results in 2 or 3. Find the numerical values of the following.
a) pX(2.5)=
b) pX(1)=
a) 0
b) 80/243
a) Well, you can't really roll the die 2.5 times, so that probability is going to be zero. Unless you have a really bouncy die that goes halfway between results. In that case, good luck!
b) Okay, so we're looking for the probability of rolling a 2 or 3 exactly once. Let's think about it. Out of the 6 possible outcomes, there are two favorable outcomes - rolling a 2 or a 3. So the probability of rolling a 2 or 3 on one roll is 2/6, which simplifies to 1/3.
Since we're rolling the die 5 times independently, the probability of rolling a 2 or 3 exactly once on one roll is still 1/3. Therefore, the probability of rolling a 2 or 3 exactly once in 5 rolls is 1/3 * 1/3 * 2/3 * 2/3 * 2/3.
Calculating that delightful expression, we find that pX(1) = 8/243.
To find the numerical values of pX(2.5) and pX(1), we need to determine the probabilities of getting a certain number of 2's or 3's when rolling a fair six-sided die 5 times.
a) pX(2.5):
Since X is defined as the number of times the roll results in 2 or 3, it can only take whole number values. Therefore, pX(2.5) is not a valid probability.
b) pX(1):
To calculate pX(1), we need to determine the probability of getting exactly 1 2 or 3 when rolling a fair six-sided die 5 times.
The probability of rolling a 2 or 3 on a single roll is 2/6 = 1/3, as there are two favorable outcomes (rolling a 2 or 3) out of six possible outcomes.
Now, to find the probability of getting exactly 1 2 or 3 in 5 rolls, we can use the binomial probability formula:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
where:
- n is the number of trials (5 rolls in this case)
- k is the number of successful outcomes (1 2 or 3 in this case)
- p is the probability of a successful outcome (1/3 in this case)
- (n choose k) is the binomial coefficient, given by (n!)/(k!(n-k)!)
Using the formula, we can calculate pX(1) as follows:
pX(1) = (5 choose 1) * (1/3)^1 * (1 - 1/3)^(5-1)
= 5 * (1/3) * (2/3)^4
= 5 * 1/3 * 16/81
= 80/243
Therefore, the numerical value of pX(1) is 80/243.
To find the numerical values of the probabilities, we need to determine the probability mass function (PMF) of random variable X, which represents the number of times the roll results in 2 or 3.
Let's break down the problem step by step.
Step 1: Determine the possible values of X.
In this case, X can take values from 0 to 5, as it represents the number of times the roll results in 2 or 3 in 5 independent trials.
Step 2: Calculate the probability of each value of X.
To calculate the probability of X = 0, we need to find the probability that none of the rolls results in 2 or 3. Since the die has 6 possible outcomes, and we want to avoid two specific outcomes (2 and 3), the probability of any single trial not resulting in a 2 or 3 is 4/6 or 2/3. As the trials are independent, we can multiply these probabilities together to find the probability of none of the rolls being 2 or 3:
P(X = 0) = (2/3)^5 ≈ 0.131
To calculate the probability of X = 1, we need to find the probability that exactly one of the rolls results in 2 or 3. There are 5 possible rolls that can be the "success" (2 or 3), and for each of them, the probability is 2/6 or 1/3. The remaining 4 rolls must not be 2 or 3, so the probability is 4/6 or 2/3. We multiply these probabilities together:
P(X = 1) = 5 * (1/3) * (2/3)^4 ≈ 0.329
Similarly, we can calculate the probabilities for X = 2, X = 3, X = 4, and X = 5:
P(X = 2) = (5 choose 2) * (1/3)^2 * (2/3)^3 ≈ 0.329
P(X = 3) = (5 choose 3) * (1/3)^3 * (2/3)^2 ≈ 0.164
P(X = 4) = (5 choose 4) * (1/3)^4 * (2/3)^1 ≈ 0.055
P(X = 5) = (5 choose 5) * (1/3)^5 * (2/3)^0 ≈ 0.008
Step 3: Calculate the numerical values of the probabilities.
Now let's find the numerical values of the two probabilities mentioned in the question:
a) pX(2.5) refers to the probability that X takes the value 2.5. Since X can only take integer values, pX(2.5) = 0.
b) pX(1) refers to the probability that X takes the value 1. We have already calculated this probability:
pX(1) ≈ 0.329
Therefore,
a) pX(2.5) = 0
b) pX(1) ≈ 0.329