Question

1. calculate the probabilities of each event happening:

a) rolling a sum less than 6 with two die.
b) rolling a sum less than 6 with two die,
if you knew the first die rolled was a 4
c) explain how including the condition changes the probability of rolling a sum less than 6.

Explanation:

Step1: Calculate total outcomes for two - die roll

When rolling two dice, the total number of outcomes is \(n(S)=6\times6 = 36\)

Step2: Find favorable outcomes for sum less than 6

The possible sums less than 6:

• Sum = 2: \((1,1)\) (1 outcome)
• Sum = 3: \((1,2),(2,1)\) (2 outcomes)
• Sum = 4: \((1,3),(2,2),(3,1)\) (3 outcomes)
• Sum = 5: \((1,4),(2,3),(3,2),(4,1)\) (4 outcomes)

Total favorable outcomes \(n(A)=1 + 2+3 + 4=10\)
Probability \(P(A)=\frac{n(A)}{n(S)}=\frac{10}{36}=\frac{5}{18}\approx0.2778\)

Step3: Conditional probability for part b

If the first die is 4, then the second die can be 1, 2, 3, 4, 5, 6. Total \(n = 6\)
For sum less than 6 (sum = 4 + x<6, so x < 2). When first die is 4, the second die must be 1. So \(n(B)=1\)
Probability \(P(B)=\frac{1}{6}\approx0.1667\)

Step4: Explain the change in probability

In part a), we consider all possible pairs of two - dice rolls. The sample space is \(n(S)=36\). In part b), we have a condition (first die is 4). This reduces the sample space. Instead of considering all 36 possible outcomes of two - dice rolls, we only consider the 6 possible outcomes where the first die is fixed at 4. So the conditional probability (part b) is calculated with a smaller sample space compared to the unconditional probability (part a)

Answer:

a) \(\frac{5}{18}\approx0.2778\)
b) \(\frac{1}{6}\approx0.1667\)
c) In part a), the probability is calculated over the entire sample space of two - dice rolls (\(n = 36\)). In part b), the condition (first die is 4) reduces the sample space to \(n = 6\). Since the sample space is smaller in part b), the probability value changes.