Question

3 dice are tossed. Find the probability that sum of digits is 14 (a) \(\left(21 / 6^{3}\right)\) (b) \(\left(15 / 6^{3}\right)\) (c) \(\left(27 / 6^{3}\right)\) (d) \(\left(16 / 6^{3}\right)\)

Short answer

The probability that the sum of the digits is 14 when 3 dice are tossed is (b) \(\left(15 / 6^{3}\right)\).

Step-by-step solution

Determine the total number of possible outcomes

When 3 dice are rolled, each die has 6 possible outcomes (since there are 6 faces on each die). To find the total number of possible outcomes, we use the multiplication rule, which states that the number of possible outcomes in a sequence of independent events can be calculated by multiplying together the number of possibilities for each event. In this case, the total number of possible outcomes is \(6^3\).

Determine the number of successful outcomes

In order to find the number of successful outcomes, i.e., the total number of outcomes where the sum of the numbers on the dice is equal to 14, we can consider different possible cases: 1. One die shows a 6, the second die shows a 6, and the third die shows a 2. There are 3! = 6 ways to arrange this case. 2. One die shows a 6, the second die shows a 5, and the third die shows a 3. There are 3! = 6 ways to arrange this case. 3. One die shows a 5, the second die shows a 5, and the third die shows a 4. There are 3!/2! = 3 ways to arrange this case (since we have two repeating numbers). So, the total number of successful outcomes is 6 + 6 + 3 = 15.

Calculate the probability

Now that we have the total number of possible outcomes (\(6^3 = 216\)) and the total number of successful outcomes (15), we can divide the number of successful outcomes by the total number of possible outcomes to find the probability: Probability = \(\frac{\text{Number of successful outcomes}}{\text{Total number of possible outcomes}} = \frac{15}{216}\) Therefore, the probability that the sum of the digits is 14 when 3 dice are tossed is: \[\frac{15}{6^3}\] Comparing the result to the given options, we can see that the answer is (b) \(\left(15 / 6^{3}\right)\).

Key concepts

Combinatorial Probability

Combinatorial probability is all about finding the likelihood of certain outcomes using combinations. When dealing with dice, each roll has multiple outcomes that can combine in different ways. By using combinations, we can determine how many specific arrangements are possible for a desired outcome. This is key for problems involving multiple dice rolls because each die operates independently. So, the main task is to figure out how the combinations of numbers can achieve a particular sum. By analyzing combinations, we identify not just what is possible, but also how probable each scenario is.

Dice Outcomes

When rolling multiple dice, each die can land in one of six possible ways since it has six faces. This means that for 3 dice, there are a total of

• 6 potential outcomes per die
• 3 dice
• The total outcomes equal to \(6^3 = 216\)

Each combination of three numbers is a sequence consisting of these dice outcomes. Understanding these outcomes is critical, as it forms the basis of probability calculations. We know there can be many permutations of dice leading to various sums, yet we focus on how these sequences combine to reach a specific total, like the number 14 in this problem.

Probability Calculation

Calculating probability involves comparing successful outcomes to all possible outcomes. Here, when seeking the probability of a sum of 14, we need to:

• Identify successful outcomes (where dice total 14)
• Calculate the number of such successful outcomes
• Divide by the total number of possible outcomes, \(216\)

So, for this problem, with 15 successful combinations (like combinations of 6, 6, 2 or 6, 5, 3), the probability is \(\frac{15}{216}\). This calculation explains how probable it is to roll a sum of 14 with three dice, leveraging both combination insights and outcome counting.

Sum of Dice Outcomes

Finding the sum of dice outcomes requires understanding how different face values contribute to a particular sum. For our scenario, the target was a sum of 14. This requires thinking about which combinations of six-face numbers add up to 14:

• Examples include combinations like 6 + 6 + 2 or 5 + 5 + 4.
• Each valid combination must be counted in all possible permutations
• The key is ensuring repetitions (like two fives) are accounted for without duplication

Arranging these values in all their possible ways gives us successful outcomes, crucial to determining the probability. Understanding how to achieve a specific sum is essential and connects directly to the basics of combinatorial probability.