Math Problem Statement

Trevion has a bag of marbles with 1 blue marbles, 1 white marbles, and 2 red marbles.

Find the following probabilities of Trevion drawing the given marbles from the bag if the first marble(s) is(are) returned to the bag after they are drawn.

a) A Blue, then a red =

b) A red, then a white =

c) A Blue, then a Blue, then a Blue =

Solution

We will calculate each probability step by step based on the number of marbles in Trevion's bag and the rule that the first marble is returned to the bag after each draw (which means each draw is independent).

Trevion has:

  • 1 blue marble
  • 1 white marble
  • 2 red marbles

This gives a total of 1+1+2=41 + 1 + 2 = 4 marbles.

a) Probability of drawing a blue marble, then a red marble:

The probability of drawing a blue marble on the first draw is: P(Blue)=14P(\text{Blue}) = \frac{1}{4}

The probability of drawing a red marble on the second draw (after the first marble is returned) is: P(Red)=24=12P(\text{Red}) = \frac{2}{4} = \frac{1}{2}

Since the events are independent, the total probability is: P(Blue, then Red)=P(Blue)×P(Red)=14×12=18P(\text{Blue, then Red}) = P(\text{Blue}) \times P(\text{Red}) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}

b) Probability of drawing a red marble, then a white marble:

The probability of drawing a red marble on the first draw is: P(Red)=24=12P(\text{Red}) = \frac{2}{4} = \frac{1}{2}

The probability of drawing a white marble on the second draw is: P(White)=14P(\text{White}) = \frac{1}{4}

Again, since the events are independent, the total probability is: P(Red, then White)=P(Red)×P(White)=12×14=18P(\text{Red, then White}) = P(\text{Red}) \times P(\text{White}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}

c) Probability of drawing a blue marble three times in a row:

Since the marble is returned after each draw, the probability of drawing a blue marble is the same for each draw: P(Blue)=14P(\text{Blue}) = \frac{1}{4}

Thus, the total probability of drawing a blue marble three times is: P(Blue, Blue, Blue)=P(Blue)×P(Blue)×P(Blue)=(14)3=164P(\text{Blue, Blue, Blue}) = P(\text{Blue}) \times P(\text{Blue}) \times P(\text{Blue}) = \left(\frac{1}{4}\right)^3 = \frac{1}{64}

Final answers:

a) P(Blue, then Red)=18P(\text{Blue, then Red}) = \frac{1}{8}

b) P(Red, then White)=18P(\text{Red, then White}) = \frac{1}{8}

c) P(Blue, Blue, Blue)=164P(\text{Blue, Blue, Blue}) = \frac{1}{64}


Would you like more details or have any questions about these calculations?

Here are 5 related questions:

  1. What is the probability of drawing two white marbles in a row?
  2. What is the probability of drawing a red marble first, then a blue marble?
  3. What is the probability of drawing a red marble three times in a row?
  4. If Trevion adds one more red marble, how would the probabilities change?
  5. What is the probability of drawing one marble of each color in three draws?

Tip: When events are independent, you can find the total probability by multiplying the probabilities of each event.

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Math Problem Analysis

Mathematical Concepts

Probability
Independent Events

Formulas

P(A and B) = P(A) * P(B) (for independent events)
Probability of an event = Number of favorable outcomes / Total outcomes

Theorems

Multiplication Rule of Probability for Independent Events

Suitable Grade Level

Grades 6-8