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A fair die is a cube with number 1 through 6 on the sides , represented as painted spots. If a fair die is rolled, what is the probability of number 6 landing face up? 1/6 COMPLETE With a fair die, the probability of rolling any number 1 through 6 is the same. If P(1) re epresents the probability of rolling a 1, P(2)the probability of rolling a 2, and so forth, what is the value of: P(1)+P(2)+P(3)+P(4)+ P(5)+P(6) square

Pergunta

A fair die is a cube with number 1 through 6 on
the sides , represented as painted spots. If a fair
die is rolled, what is the probability of number 6
landing face up?
1/6
COMPLETE
With a fair die, the probability of rolling any
number 1 through 6 is the same.
If P(1) re epresents the probability of rolling a 1,
P(2)the probability of rolling a 2, and so forth,
what is the value of: P(1)+P(2)+P(3)+P(4)+
P(5)+P(6) square

A fair die is a cube with number 1 through 6 on the sides , represented as painted spots. If a fair die is rolled, what is the probability of number 6 landing face up? 1/6 COMPLETE With a fair die, the probability of rolling any number 1 through 6 is the same. If P(1) re epresents the probability of rolling a 1, P(2)the probability of rolling a 2, and so forth, what is the value of: P(1)+P(2)+P(3)+P(4)+ P(5)+P(6) square

Solução

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LucianoProfissional · Tutor por 6 anos

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The sum of the probabilities of rolling each number on a fair die is 1.

Explicação

## Step 1<br />The problem involves the concept of probability. A fair die is a cube with numbers 1 through 6 on its sides. Each time the die is rolled, each number has an equal chance of appearing. This means that the probability of rolling any number from 1 to 6 is the same.<br /><br />## Step 2<br />The probability of rolling a specific number, such as 1, 2, 3, 4, 5, or 6, is \( \frac{1}{6} \). This is because there are six possible outcomes when rolling a fair die, and each outcome is equally likely.<br /><br />## Step 3<br />The problem asks for the sum of the probabilities of rolling each number. This can be calculated by adding the individual probabilities together.<br /><br />### \( P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \)<br /><br />## Step 4<br />When these fractions are added together, they sum up to 1. This is because the total probability of all possible outcomes is always 1.
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