Question

The probability that Eric drives to work is 0.28 and the probability that he brings a sandwich for lunch is 0.36. The probability that he drives to work and brings a sandwich for lunch is 0.18. What is the probability that he brings a sandwich to work for lunch given that he drives there?

Answer 1

To find the probability that Eric brings a sandwich to work for lunch given that he drives there, you need to use the concept of conditional probability.

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 given that B has occurred, \( P(A \cap B) \) is the probability of both A and B occurring, and \( P(B) \) is the probability of event B.

In this question, let A be the event that Eric brings a sandwich for lunch, and B be the event that he drives to work.

Given:
- \( P(B) = P(\text{Drive}) = 0.28 \)
- \( P(A) = P(\text{Sandwich}) = 0.36 \) (though we do not use this value for this particular computation)
- \( P(A \cap B) = P(\text{Drive and Sandwich}) = 0.18 \)

Now using the formula for conditional probability, plug in the values for A and B:

\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]

\[ P(\textSandwich) = \frac{P(\text{Drive and Sandwich})}{P(\text{Drive})} \]

\[ P(\textDrive) = \frac{0.18}{0.28} \]

Now, we can calculate the value:

\[ P(\textDrive) = \frac{18}{28} \]

If we simplify the fraction by dividing both numerator and denominator by 2, we get:

\[ P(\textSandwich) = \frac{9}{14} \]

To get a decimal value, we divide 9 by 14:

\[ P(\textDrive) \approx 0.642857142857143 \]

So the probability that Eric brings a sandwich to work for lunch given that he drives there is approximately 0.643, or 64.3% when expressed as a percentage.

Answer 2

The probability that Eric brings a sandwich to work for lunch given that he drives there is 0.64.

Step-by-step explanation:

To find the conditional probability that Eric brings a sandwich for lunch given that he drives to work, we can use the formula for conditional probability:

`\[ P(\text{Sandwich} | \text{Drives})`=
`\frac{P(\text{Sandwich} \cap \text{Drives})}{P(\text{Drives})} \]`

Given that
`\(P(\text{Drives})` =
`0.28\), \(P(\text{Sandwich})` = 0.36\), and
`\(P(\text{Sandwich} \cap \text{Drives})`= 0.18\), we can substitute these values into the formula:

`\[ P(\text{Sandwich} | \text{Drives})`=
`(0.18)/(0.28)` =
`0.64 \]`

Therefore, the probability that Eric brings a sandwich to work for lunch given that he drives there is 0.64.

This problem involves conditional probability, which is the likelihood of an event occurring given that another event has already occurred. We are given the probabilities of Eric driving to work (0.28), bringing a sandwich for lunch (0.36), and the probability of both events happening together (0.18).

To find the probability that he brings a sandwich given that he drives, we use the formula for conditional probability:
`\( P(\text{A} | \text{B})` =
`\frac{P(\text{A} \cap \text{B})}{P(\text{B})} \)`. Substituting the given values, we get
`\( P(\text{Sandwich} | \text{Drives})` = \f
`rac{0.18}{0.28}` =
`0.64 \)`.

Conditional probability calculates the chance of an event happening given that another event has already occurred. In this case, given that Eric drives to work, we're evaluating the probability of him bringing a sandwich. The probability is significantly affected by the intersection of the two events (driving and bringing a sandwich) compared to the overall probability of driving.

In this scenario, the likelihood of Eric bringing a sandwich to work when he drives is notably higher (0.64 or 64%) than the standalone probability of bringing a sandwich (0.36 or 36%), showcasing the relationship between these events.