Dice Probability Calculation
The problem involves finding the probability of a specific range for the sum ($X$) of outcomes when rolling two fair dice. The required probability is for the sum being 'at least 4' and 'at most 8', which mathematically translates to $P(4 \le X \le 8)$.
Two Dice Outcomes
A standard fair die has 6 faces (numbered 1 to 6). When rolling two fair dice, the total number of possible outcomes is $6 \times 6 = 36$. Each outcome is equally likely. The random variable $X$ represents the sum of the numbers shown on the two dice.
Probability for Sums 4 to 8
To find $P(4 \le X \le 8)$, we need to count the number of outcomes where the sum $X$ is 4, 5, 6, 7, or 8.
- Sum = 4: The pairs are (1,3), (2,2), (3,1). There are 3 outcomes.
- Sum = 5: The pairs are (1,4), (2,3), (3,2), (4,1). There are 4 outcomes.
- Sum = 6: The pairs are (1,5), (2,4), (3,3), (4,2), (5,1). There are 5 outcomes.
- Sum = 7: The pairs are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). There are 6 outcomes.
- Sum = 8: The pairs are (2,6), (3,5), (4,4), (5,3), (6,2). There are 5 outcomes.
The total number of favorable outcomes is the sum of these counts: $3 + 4 + 5 + 6 + 5 = 23$.
Final Probability Result
The probability is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes.
$ P(4 \le X \le 8) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} $ $ P(4 \le X \le 8) = \frac{23}{36} $
Thus, the probability that the sum of the two dice is at least 4 and at most 8 is 23/36.
Important Questions from Discrete Distributions
The value of a and b so that the following is probability mass function
X: 0 1 2 P(X = x): 3a 3b 4b with mean 1.1, is:
Digital data received from a sensor can fill up 0 to 32 buffers. Let the sample space be
S = {0, 1, 2, .........., 32} where the sample j denote that j of the buffers are full and \(p\left( i \right) = \frac{1}{{561}}\left( {33 - i} \right)\)
. Let A denote the event that the even number of buffers are full. Then p(A) is :If X is a Poisson random variate with mean 3, then P(|X- 3| < 1) will be:
Let x ∼ N(μ, σ2) If μ2 = σ2, (μ > 0), then the value of P(X < -μ | X < μ) in terms of cumulative function N (0, 1) is:
Consider a binomial random variable X. If X1, X2,...Xn are independent and identically distributed samples from the distribution of X with sum \(Y = \mathop \sum \limits_{i = 1}^n {X_i}\) then the distribution of Y as n → ∞ can be approximated as.
