Question

The probabilities of different faces of biased dice to appear are as follows Face number \(1 \quad 2\) \(\begin{array}{llll}3 & 4 & 5\end{array}\) Probability \(\begin{array}{llllll}0.1 & 0.32 & 0.21 & 0.15 & 0.05 & 0.17\end{array}\) The dice is thrown and it is known that either the face number 1 or 2 will appear. Then, the probability of the face number 1 to appear is (1) \(5 / 21\) (2) \(5 / 12\) (3) \(7 / 23\) (4) \(3 / 10\)

Short answer

(1) \( \frac{5}{21} \)

Step-by-step solution

- Identify the given probabilities

The probability distribution of the biased dice is provided. The probabilities for each face of the dice are: Face 1: 0.1Face 2: 0.32Face 3: 0.21Face 4: 0.15Face 5: 0.05Face 6: 0.17

- Understand the condition

It is known that either face number 1 or face number 2 will appear. This means that the sample space is restricted to only face 1 and face 2. The total probability in this restricted sample space is: \[ P(1) + P(2) = 0.1 + 0.32 = 0.42 \]

- Use conditional probability

The problem requires finding the probability of face number 1 appearing given that either face 1 or face 2 appears. According to conditional probability, we have: \[ P(1 \text{ appears} \,| \text{1 or 2 appears}) = \frac{P(1)}{P(1) + P(2)} \]

- Substitute the values

Substitute the known probabilities into the conditional probability formula: \[ P(1 \text{ appears} \,| \text{1 or 2 appears}) = \frac{0.1}{0.1 + 0.32} = \frac{0.1}{0.42} = \frac{10}{42} = \frac{5}{21} \]

- Choose the correct option

Comparing the calculated probability with the given options, the correct answer is: (1) \( \frac{5}{21} \)

Key concepts

biased dice

A biased dice is not like a fair dice where each face has an equal probability of landing face up. Instead, certain faces are more likely to appear than others. In this exercise, the dice is biased, meaning its faces have different probabilities. For example, the probability for face 1 to appear is 0.1, while for face 2 it’s 0.32. These varying probabilities are what we call the probability distribution of the biased dice.

probability distribution

A probability distribution lists all possible outcomes and their corresponding probabilities. Each value in the distribution represents the likelihood of a particular face appearing when the dice is rolled. In this exercise, our distribution for the dice faces is:

• Face 1: 0.1
• Face 2: 0.32
• Face 3: 0.21
• Face 4: 0.15
• Face 5: 0.05
• Face 6: 0.17

The sum of all probabilities in a distribution must equal 1. This helps ensure that every outcome is accounted for and makes it easier for us to work with the probabilities.

restricted sample space

A restricted sample space means narrowing down the possible outcomes to a specific subset dictated by a condition. In this problem, we are told that either face 1 or face 2 will appear. This restriction modifies our sample space. Our total probability now only includes face 1 and face 2. The sum of these probabilities is 0.1 for face 1 plus 0.32 for face 2, totaling 0.42. This restricted sample space helps us focus on the relevant outcome probabilities while ignoring the other faces.

conditional probability method

The conditional probability method calculates the probability of an event occurring given that another event has already occurred. The formula for conditional probability is: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \], where \ P(A|B) \ is the probability of event A occurring given event B has occurred. In our problem, we need to find the probability of face 1 showing up given that either face 1 or face 2 shows up. This is calculated using: \[ P(1 \text{ appears} | \text{1 or 2 appears}) = \frac{P(1)}{P(1) + P(2)} \]. Substituting the values, we get: \[ P(1 | 1 \text{ or } 2 ) = \frac{0.1}{0.1 + 0.32} = \frac{0.1}{0.42} = \frac{5}{21} \]. Understanding this method allows us to draw meaningful insights from situations where only specific outcomes are considered.