Question

Find the probability of throwing at least one of the following totals on a single throw of a pair of dice: a total of 5, a total of 6 , or a total of 7 . Define the events \(A, B\), and \(C\) as follows: Event \(\mathrm{A}\) : a total of 5 is thrown, Event \(B\) : a total of 6 is thrown, Event \(C\) : a total of 7 is thrown.

Short answer

The probability of throwing a total of 5, 6, or 7 with a single throw of a pair of dice is \(\frac{5}{12}\).

Step-by-step solution

Calculate the independent probabilities

To find the probability of each event, we need to compute the number of favorable outcomes for each total and divide by the total number of possible outcomes. In this case, the total number of possible outcomes is 36 because there are 6 sides on each dice, and we have two dice, so there are 6 x 6 = 36 possible outcomes. Event A: A total of 5 is thrown There are four possible ways to get a total of 5: (1, 4), (2, 3), (3, 2), (4, 1). Thus, we have 4 favorable outcomes out of 36 possible outcomes. P(A) = \(\frac{4}{36}\) Event B: A total of 6 is thrown There are five possible ways to get a total of 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1). Thus, we have 5 favorable outcomes out of 36 possible outcomes. P(B) = \(\frac{5}{36}\) Event C: A total of 7 is thrown There are six possible ways to get a total of 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). Thus, we have 6 favorable outcomes out of 36 possible outcomes. P(C) = \(\frac{6}{36}\)

Calculate the unions of the independent probabilities

Since events A, B, and C are mutually exclusive (i.e., no two events can occur at the same time), we can find the union of these events by simply adding their probabilities. P(A ∪ B ∪ C) = P(A) + P(B) + P(C)

Calculate the probability of the union of the events

Now we plug in the values calculated in Step 1 and find the probability of the union of the events. P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = \(\frac{4}{36}+\frac{5}{36}+\frac{6}{36}\) P(A ∪ B ∪ C) = \(\frac{15}{36}\)

Simplify the probability

Finally, we simplify the probability by dividing both the numerator and the denominator by the greatest common divisor, which is 3 in this case. P(A ∪ B ∪ C) = \(\frac{15}{36}\) = \(\frac{15 \div 3}{36 \div 3}\) = \(\frac{5}{12}\) So the probability of throwing a total of 5, 6, or 7 with a single throw of a pair of dice is \(\frac{5}{12}\).

Key concepts

Independent Probabilities

When we roll a pair of dice, it's like watching two separate little universes where each die does its own thing, unaware of the other's existence. This is the core idea behind independent probabilities—the outcome of one event doesn't influence the outcome of the other. If we're curious about the chances of both dice coming up with certain numbers, we simply multiply their independent probabilities together.

Why bother with independent probabilities? It's because they give us the power to measure how likely two or more events are to occur together, without one messing with the other. When we calculated the probabilities of getting specific totals from two dice, we counted each combination that can give us those totals, taking into account the independence of each die.

Mutually Exclusive Events

Let's talk about events that can cause a stir when they try to show up at the same party. Mutually exclusive events are like two people who can't stand each other—they simply won't occur at the same time. In our dice-rolling scenario, if we throw a total of 5, we certainly can't throw a 6 or 7 on the same roll; it's a one-or-the-other deal. That's mutual exclusivity in action!

Understanding this gives us a neat trick for finding the probability of 'at least one' of these events happening. Since they're all giving each other the cold shoulder (being mutually exclusive), we just add their probabilities up to find the chance of any one of them happening. This concept is essential because it lets us neatly add up probabilities without worrying about overlapping outcomes, which would be a no-no for events that don't like to share their moment.

Favorable Outcomes

Here's the spotlight moment: favorable outcomes. Imagine if you had a bag full of different candies, and you wanted only the chocolate ones. Those chocolates are your 'favorable outcomes.' With dice, we considered the different ways we could roll certain totals and called those our favorable outcomes. Why? Because they're what we're rooting for.

Diving into the sweetness of probability, if you've got a fistful of possibilities but you only care about a few, those few are your golden tickets—the favorable outcomes. Figuring out these precious few amidst the chaos of possibility lets us peek into the future and guess our chances of success. If you're committed to understanding just one thing in probability, grab onto the concept of favorable outcomes like it's your last chocolate on Halloween night.