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Chapter 7: Problem 48

If Suzan grabs two marbles, one at a time, out of a bag of five red marbles and four green ones, find an event with a probability that depends on the order in which the two marbles are drawn.

Short Answer

Expert verified
The probability of the event happening in these specific orders is \(\dfrac{40}{72}\) or simplified: \(\dfrac{5}{9}\).

Step by step solution

01

List the possibilities of grabbing two marbles in order

There are two possibilities when drawing two marbles - one at a time - from the bag: 1. Drawing a red marble first and then drawing a green marble. 2. Drawing a green marble first and then drawing a red marble.
02

Calculate the probability of drawing a red marble first and then a green marble

We first calculate the probability of drawing a red marble. Since there are 5 red marbles and 9 marbles in total, the probability is: \(P(\text{Red}_{1}) = \dfrac{5}{9}\) After drawing a red marble, there are now 8 marbles left in the bag (4 green and 4 red). The probability of drawing a green marble in the second draw is: \(P(\text{Green}_{2}|\text{Red}_{1}) = \dfrac{4}{8}\) Now, we multiply the probabilities to get the probability of this event occurring in this specific order: \(P(\text{Red}_{1}\text{ and } \text{Green}_{2}) = P(\text{Red}_{1}) \cdot P(\text{Green}_{2} | \text{Red}_{1}) = \dfrac{5}{9} \cdot \dfrac{4}{8} = \dfrac{20}{72}\)
03

Calculate the probability of drawing a green marble first and then a red marble

First, calculate the probability of drawing a green marble. There are 4 green marbles and 9 marbles in total, so the probability is: \(P(\text{Green}_{1}) = \dfrac{4}{9}\) After drawing a green marble, there are now 8 marbles left in the bag (5 red and 3 green). The probability of drawing a red marble in the second draw is: \(P(\text{Red}_{2} | \text{Green}_{1}) = \dfrac{5}{8}\) Now, we multiply the probabilities to get the probability of this event occurring in this specific order: \(P(\text{Green}_{1}\text{ and } \text{Red}_{2}) = P(\text{Green}_{1}) \cdot P(\text{Red}_{2} | \text{Green}_{1}) = \dfrac{4}{9} \cdot \dfrac{5}{8} = \dfrac{20}{72}\)
04

Combine the probabilities of the two events

The event we are looking for is either drawing a red marble first and then a green marble, or drawing a green marble first and then a red marble. This means the probabilities of these two ordered events should be added together: \(P(\text{Event}) = P(\text{Red}_{1}\text{ and } \text{Green}_{2}) + P(\text{Green}_{1}\text{ and } \text{Red}_{2}) = \dfrac{20}{72} + \dfrac{20}{72} = \dfrac{40}{72}\) So, the probability of the event happening in these specific orders is \(\dfrac{40}{72}\) or simplified: \(\dfrac{5}{9}\).

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