Question

1. a bag contains 4 blue marbles, 5 yellow marbles, and 3 red marbles. if you draw one marble at random, find the probability of yellow or blue.

a) \\( \frac { 3 } { 4 } \\) b) \\( \frac { 5 } { 12 } \\)
c) \\( \frac { 7 } { 12 } \\) d) \\( \frac { 2 } { 3 } \\)

1. a bag contains 4 blue marbles, 5 yellow marbles, and 3 red marbles. if you draw one marble at random, find the probability of not yellow.

a) \\( \frac { 5 } { 12 } \\) b) \\( \frac { 7 } { 12 } \\)
c) \\( \frac { 1 } { 4 } \\) d) \\( \frac { 3 } { 4 } \\)

1. if you draw a marble, replace it, and draw again from the same bag (4 blue, 5 yellow, 3 red), find the probability of red then yellow.

a) \\( \frac { 5 } { 48 } \\) b) \\( \frac { 1 } { 9 } \\)
c) \\( \frac { 1 } { 11 } \\) d) \\( \frac { 5 } { 36 } \\)

1. if you draw a marble, replace it, and draw again from the same bag (4 blue, 5 yellow, 3 red), find the probability of both blue.

a) \\( \frac { 1 } { 9 } \\) b) \\( \frac { 1 } { 16 } \\)
c) \\( \frac { 4 } { 33 } \\) d) \\( \frac { 5 } { 36 } \\)

1. if you draw a marble, do not replace it, and draw again from the same bag (4 blue, 5 yellow, 3 red), find the probability of red then blue.

a) \\( \frac { 5 } { 33 } \\) b) \\( \frac { 5 } { 36 } \\)
c) \\( \frac { 1 } { 9 } \\) d) \\( \frac { 1 } { 11 } \\)

Explanation:

Step1: Calculate total number of marbles

Total marbles = \(4 + 5+3=12\)

Step2: Calculate probability of yellow or blue

Number of yellow or blue marbles = \(4 + 5 = 9\)
Probability \(P=\frac{9}{12}=\frac{3}{4}\)

Step3: Calculate probability of not yellow

Number of non - yellow marbles = \(4 + 3=7\)
Probability \(P=\frac{7}{12}\)

Step4: Calculate probability of red then yellow (with replacement)

Probability of red \(P_{1}=\frac{3}{12}\), probability of yellow \(P_{2}=\frac{5}{12}\)
Since events are independent (with replacement), \(P = P_{1}\times P_{2}=\frac{3}{12}\times\frac{5}{12}=\frac{5}{48}\)

Step5: Calculate probability of both blue (with replacement)

Probability of blue first \(P_{1}=\frac{4}{12}\), probability of blue second \(P_{2}=\frac{4}{12}\)
Since events are independent (with replacement), \(P=\frac{4}{12}\times\frac{4}{12}=\frac{1}{9}\)

Step6: Calculate probability of red then blue (without replacement)

Probability of red first \(P_{1}=\frac{3}{12}\), then number of marbles left = \(11\)
Probability of blue second \(P_{2}=\frac{4}{11}\)
Since events are dependent (without replacement), \(P=\frac{3}{12}\times\frac{4}{11}=\frac{1}{11}\)

Answer:

1. a) \(\frac{3}{4}\)
2. b) \(\frac{7}{12}\)
3. a) \(\frac{5}{48}\)
4. a) \(\frac{1}{9}\)
5. d) \(\frac{1}{11}\)