If P(A)=59 and P(B)=29 and the two events are independent, what is P(A and B)?(1 point)

Responses

109
Start Fraction 10 over 9 End Fraction

1081
Start Fraction 10 over 81 End Fraction

79
Start Fraction 7 over 9 End Fraction

718
Start Fraction 7 over 18 End Fraction

1 answer

To find \( P(A \text{ and } B) \) for two independent events \( A \) and \( B \), we use the formula:

\[ P(A \text{ and } B) = P(A) \times P(B) \]

In this case, \( P(A) = 59 \) and \( P(B) = 29 \). However, these values represent percentages, so we need to convert them to decimal probabilities:

\[ P(A) = \frac{59}{100} \quad \text{and} \quad P(B) = \frac{29}{100} \]

Now, we can calculate \( P(A \text{ and } B) \):

\[ P(A \text{ and } B) = \frac{59}{100} \times \frac{29}{100} = \frac{59 \times 29}{100 \times 100} = \frac{1711}{10000} \]

Now let's simplify \( \frac{1711}{10000} \) or convert it to a proper fraction or decimal if necessary:

  1. \( 1711 \) is the product of \( 59 \) and \( 29 \). It's already in its simplest form.
  2. The fraction \( \frac{1711}{10000} \) does not correspond to any of the options provided directly.

Since none of the provided responses match \( P(A \text{ and } B) = 1711/10000 \), let's double-check the breakdown.

To match the provided options, we realize they are represented in a different format:

  • Option 1: 109 is clearly incorrect.
  • Option 2: Start Fraction 10 over 9 End Fraction is also incorrect.
  • Option 3: 1081 is not the value.
  • Option 4: Start Fraction 10 over 81 End Fraction seems wrong through verification.

However, if we check calculations with numbers generated,

One approach would be \( 59 + 29 \) being incorrect for overlap consideration so the independent properties don't lead us to a correct solution.

The overall probability \( \frac{1711}{10000} \) simplifies to fall between \( 17.11 \)% hence falls out.

Thus, concluding that \( P(A \text{ and } B) \) leads options accordingly, appearing to have transcription misconnect domains.

Conclusively, the answer is not listed hence the simplified results indicating mistakes might cater penalties instead.