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One letter is selected at random from the 5 letters \(\mathrm{V}, \mathrm{W}, \mathrm{X}, \mathrm{Y},\) and \(\mathrm{Z}\), and event \(A\) is the event that the letter \(\mathrm{V}\) is selected. A fair six-sided die with sides numbered \(1,2,3,4,5,\) and 6 is to be rolled, and event \(B\) is the event that a 5 or a 6 shows. A fair coin is to be tossed, and event \(C\) is the event that a head shows. What is the probability that event \(A\) occurs and at least one of the events \(B\) and \(C\) occurs? a. \(\frac{1}{30}\) b. \(\frac{1}{15}\) c. \(\frac{1}{10}\) d. \(\frac{2}{15}\) e. \(\frac{1}{5}\)

Short Answer

Expert verified
\( \frac{2}{15} \)

Step by step solution

01

- Determine the probability of event A

Event A is choosing the letter \mathrm{V} from the set {\( \mathrm{V}, \mathrm{W}, \mathrm{X}, \mathrm{Y}, \mathrm{Z} \)}. There are 5 letters, so the probability \(P(A)\) is \( \frac{1}{5} \).
02

- Determine the probability of event B

Event B is rolling a 5 or a 6 on a fair six-sided die. There are 2 favorable outcomes (5 and 6) out of 6 possible outcomes. So the probability \(P(B)\) is \( \frac{2}{6} = \frac{1}{3} \).
03

- Determine the probability of event C

Event C is getting a head when tossing a fair coin. There are 2 possible outcomes (head or tail). So the probability \(P(C)\) is \( \frac{1}{2} \).
04

- Determine the probability of event B or event C

We use the formula for the union of two events: \( P(B \cup C) = P(B) + P(C) - P(B \cap C) \). Since events B and C are independent, \(P(B \cap C) = P(B) \cdot P(C)\). \ \( P(B \cup C) = \frac{1}{3} + \frac{1}{2} - \left( \frac{1}{3} \cdot \frac{1}{2} \right) = \frac{1}{3} + \frac{1}{2} - \frac{1}{6} = \frac{2}{6} + \frac{3}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3} \).
05

- Calculate the probability of event A and (B or C) occurring

Since the events A and (B or C) are independent, we multiply their probabilities. \ \( P(A \cap (B \cup C)) = P(A) \cdot P(B \cup C) = \frac{1}{5} \cdot \frac{2}{3} = \frac{2}{15} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

probability rules
In probability, rules help us determine the chances of various events occurring. Basic probability is calculated as the number of favorable outcomes divided by the total number of possible outcomes. This is commonly written as
\( P(A) = \frac{number \, of \, favorable \, outcomes}{total \, number \, of \, outcomes} \).

For compound events, where two or more events are combined, we use rules like the addition rule and multiplication rule.
  • The addition rule helps us find the probability of either of two events happening. If events are mutually exclusive (cannot happen at the same time), we just add their probabilities:
    \( P(A \cup B) = P(A) + P(B) \).

  • The multiplication rule is used when events are independent (the outcome of one event doesn't affect the other). Here the probability of both events occurring is the product of their probabilities:
    \( P(A \cap B) = P(A) \cdot P(B) \).

In our exercise, we used these rules to find the probability of selecting 'V', getting a 5 or 6 on a die, or landing a head on a coin toss. Understanding these basic rules is essential for tackling more complex probability problems.
independent events
Independent events are those where the outcome of one does not affect the outcome of the other. This concept simplifies calculations, as we can multiply their probabilities directly.

Imagine we are rolling a die and flipping a coin. The result of the die roll has no impact on the coin flip outcome. Events A, B, and C in our problem (choosing 'V', rolling 5 or 6, and flipping heads) are all independent.

To find the probability of two independent events both happening, we use:
\[ P(A \cap B) = P(A) \cdot P(B) \].
This was crucial in the final step of our solution where we calculated
\( P(A \cap (B \cup C)) \).

Realizing events are independent helps break down complex problems into simpler parts, managing each part separately and then combining them.
union of events
The union of events represents the scenario where at least one of the events occurs. It’s a combination of events where we count any occurrence of any involved event.

The formula for the probability of the union of two events is:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \].

For independent events, we know
\( P(A \cap B) = P(A) \cdot P(B) \).
This adjustment subtracts the overlap where both events happen together so we don’t double-count.

In our example, we used the union formula for events B (rolling 5 or 6) and C (flipping heads), because we were interested in at least one happening. This shows how understanding the union helps manage more intricate scenarios by appropriately adding and subtracting probabilities.

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Most popular questions from this chapter

Set \(X\) consists of at least 2 members and is a set of consecutive odd integers with an average (arithmetic mean) of 37. Set \(Y\) consists of at least 10 members and is also a set of consecutive odd integers with an average (arithmetic mean) of 37. Set \(Z\) consists of all of the members of both set \(X\) and set \(Y\). Which of the following statements must be true? I. The standard deviation of set \(Z\) is not equal to the standard deviation of set \(X\). II. The standard deviation of set \(Z\) is equal to the standard deviation of set \(Y\). III. The average (arithmetic mean) of set \(Z\) is 37. a. I only b. II only c. III only d. I and III e. II and III

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If \(a\) and \(b\) are integers, and \(2 a+b=17,\) then \(8 a+b\) cannot equal which of the following? a. -1 b. 33 c. 35 d. 65 e. 71

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