To find the probability distribution of the number of successes when throwing a pair of dice four times, we first need to determine the probability of getting a doublet (two dice showing the same number) in a single throw.
Calculating the Probability of a Doublet
When rolling two dice, there are a total of 6 possible doublets: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). Since there are 36 possible outcomes when rolling two dice (6 sides on the first die multiplied by 6 sides on the second die), the probability of rolling a doublet is:
- P(Doublet) = Number of Doublets / Total Outcomes
- P(Doublet) = 6 / 36 = 1/6
Defining the Random Variable
Let X be the random variable representing the number of successes (doublets) in 4 throws. Since each throw is independent, X follows a binomial distribution:
- n = 4 (number of trials)
- p = 1/6 (probability of success)
Probability Distribution Function
The probability mass function for a binomial distribution is given by:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
Where C(n, k) is the binomial coefficient, calculated as:
C(n, k) = n! / (k!(n-k)!)
Calculating Probabilities for Each Outcome
Now, we can calculate the probabilities for k = 0, 1, 2, 3, and 4 successes:
- P(X = 0) = C(4, 0) * (1/6)^0 * (5/6)^4 = 1 * 1 * (625/1296) = 625/1296
- P(X = 1) = C(4, 1) * (1/6)^1 * (5/6)^3 = 4 * (1/6) * (125/216) = 500/1296
- P(X = 2) = C(4, 2) * (1/6)^2 * (5/6)^2 = 6 * (1/36) * (25/36) = 150/1296
- P(X = 3) = C(4, 3) * (1/6)^3 * (5/6)^1 = 4 * (1/216) * (5/6) = 20/1296
- P(X = 4) = C(4, 4) * (1/6)^4 * (5/6)^0 = 1 * (1/1296) * 1 = 1/1296
Summary of the Probability Distribution
The probability distribution of the number of successes (doublets) when throwing a pair of dice four times is:
- P(X = 0) = 625/1296
- P(X = 1) = 500/1296
- P(X = 2) = 150/1296
- P(X = 3) = 20/1296
- P(X = 4) = 1/1296
This distribution provides a clear view of the likelihood of achieving different numbers of doublets in four rolls of a pair of dice.