Question

a bag of marbles contains 8 blue marbles, 6 red marbles, and 4 green marbles. find the probability of each event if you choose marbles without looking in the bag.

1. p(green)
2. p(red)
3. p(blue)
4. you pick a marble, replace it, and then pick another. find p(blue, blue).
5. you pick a marble, replace it, and then pick another. find p(red, green).
6. you pick a marble, do not replace it, and then pick another. find p(blue, blue).
7. you pick a marble, do not replace it, and then pick another. find p(red, green).
8. you pick a marble, do not replace it, and then pick another. find p(green, green).

Explanation:

Step1: Calculate total number of marbles

The total number of marbles is \(8 + 6+4=18\).

Step2: Calculate \(P(\text{green})\) (r22)

The probability of picking a green marble is the number of green marbles divided by the total number of marbles. So \(P(\text{green})=\frac{4}{18}=\frac{2}{9}\).

Step3: Calculate \(P(\text{red})\) (r23)

The probability of picking a red marble is the number of red marbles divided by the total number of marbles. So \(P(\text{red})=\frac{6}{18}=\frac{1}{3}\).

Step4: Calculate \(P(\text{blue})\) (r24)

The probability of picking a blue marble is the number of blue marbles divided by the total number of marbles. So \(P(\text{blue})=\frac{8}{18}=\frac{4}{9}\).

Step5: Calculate \(P(\text{blue, blue})\) with replacement (r25)

Since the events are independent when we replace the marble, \(P(\text{blue, blue})=P(\text{blue})\times P(\text{blue})=\frac{4}{9}\times\frac{4}{9}=\frac{16}{81}\).

Step6: Calculate \(P(\text{red, green})\) with replacement (r26)

Since the events are independent when we replace the marble, \(P(\text{red, green})=P(\text{red})\times P(\text{green})=\frac{1}{3}\times\frac{2}{9}=\frac{2}{27}\).

Step7: Calculate \(P(\text{blue, blue})\) without replacement (r27)

The probability of the first - blue is \(\frac{8}{18}\), and the probability of the second - blue (without replacement) is \(\frac{7}{17}\). So \(P(\text{blue, blue})=\frac{8}{18}\times\frac{7}{17}=\frac{28}{153}\).

Step8: Calculate \(P(\text{red, green})\) without replacement (r28)

The probability of the first - red is \(\frac{6}{18}\), and the probability of the second - green (without replacement) is \(\frac{4}{17}\). So \(P(\text{red, green})=\frac{6}{18}\times\frac{4}{17}=\frac{4}{51}\).

Step9: Calculate \(P(\text{green, green})\) without replacement (r29)

The probability of the first - green is \(\frac{4}{18}\), and the probability of the second - green (without replacement) is \(\frac{3}{17}\). So \(P(\text{green, green})=\frac{4}{18}\times\frac{3}{17}=\frac{2}{51}\).

Answer:

r22. \(\frac{2}{9}\)
r23. \(\frac{1}{3}\)
r24. \(\frac{4}{9}\)
r25. \(\frac{16}{81}\)
r26. \(\frac{2}{27}\)
r27. \(\frac{28}{153}\)
r28. \(\frac{4}{51}\)
r29. \(\frac{2}{51}\)