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The probabilities for each of the sums are:

[tex]\[
\begin{array}{llll}
P(2) = \frac{1}{36} & P(3) = \frac{1}{18} & P(4) = \frac{1}{12} & P(5) = \frac{1}{9} \\
P(6) = \frac{5}{36} & P(7) = \frac{1}{6} & P(8) = \frac{5}{36} & P(9) = \frac{1}{9} \\
P(10) = \frac{1}{12} & P(11) = \frac{1}{18} & P(12) = \frac{1}{36} &
\end{array}
\][/tex]

The probability of a sum of 7 or 11 is:

A. [tex]\(\frac{1}{36}\)[/tex]
B. [tex]\(\frac{1}{18}\)[/tex]
C. [tex]\(\frac{2}{9}\)[/tex]
D. [tex]\(\frac{1}{36}\)[/tex]

Sagot :

GinnyAnsweridn
To determine the probability of rolling a sum of either 7 or 11 with two six-sided dice, we will follow these steps:

1. Identify the given probabilities:

- The probability of rolling a sum of 7, given as [tex]\( P(7) = \frac{1}{6} \)[/tex].
- The probability of rolling a sum of 11, given as [tex]\( P(11) = \frac{1}{18} \)[/tex].

2. Calculate the total probability of rolling a sum of 7 or 11:

Since the two events (rolling a 7 and rolling an 11) are mutually exclusive (they cannot occur simultaneously because you can only get one sum per roll), we can add these probabilities together:

[tex]\[ P(\text{Sum is 7 or 11}) = P(7) + P(11) \][/tex]

3. Substitute the given probabilities into the equation:

- [tex]\( P(7) = \frac{1}{6} \approx 0.16666666666666666 \)[/tex]
- [tex]\( P(11) = \frac{1}{18} \approx 0.05555555555555555 \)[/tex]

So,

[tex]\[ P(\text{Sum is 7 or 11}) = 0.16666666666666666 + 0.05555555555555555 = 0.2222222222222222 \][/tex]

4. Summarize the findings:

- The probability of rolling a sum of 7 is [tex]\( \approx 0.16666666666666666 \)[/tex].
- The probability of rolling a sum of 11 is [tex]\( \approx 0.05555555555555555 \)[/tex].
- Therefore, the total probability of rolling a sum of 7 or 11 is [tex]\( \approx 0.2222222222222222 \)[/tex].

Thus, the probability that the sum of the numbers on two rolled dice is either 7 or 11 is approximately [tex]\( 0.2222222222222222 \)[/tex].
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