Question

A bag of contains 4 blue marbles, 6 green marbles and 3 yellow marbles. If two marbles are drawn at random from the bag, whats the probability of:

Drawing two blue marbles? Without Replacement


.097


.087


.077


.067

Answer 1

To determine the probability of drawing two blue marbles without replacement from a bag containing 4 blue marbles, 6 green marbles, and 3 yellow marbles, we first need to find the total number of marbles in the bag.

Total marbles = 4 (blue) + 6 (green) + 3 (yellow) = 13 marbles.

Now, let's calculate the probability of drawing two blue marbles without replacement.

1. The probability of drawing the first blue marble:
\[
P(\text{First blue}) = \frac{\text{Number of blue marbles}}{\text{Total number of marbles}} = \frac{4}{13}
\]

2. After drawing the first blue marble, there are now 3 blue marbles left and a total of 12 marbles remaining in the bag. The probability of drawing the second blue marble is:
\[
P(\text{Second blue | First blue}) = \frac{\text{Number of remaining blue marbles}}{\text{Total remaining marbles}} = \frac{3}{12} = \frac{1}{4}
\]

3. Now, we can find the combined probability of both events occurring (drawing two blue marbles):
\[
P(\text{Two blue}) = P(\text{First blue}) \times P(\text{Second blue | First blue}) = \frac{4}{13} \times \frac{1}{4} = \frac{1}{13}
\]

4. Converting \(\frac{1}{13}\) into decimal form:
\[
\frac{1}{13} \approx 0.0769
\]

Now, let's round this to three decimal places. The result is approximately \(0.077\).

Thus, the probability of drawing two blue marbles without replacement is approximately **0.077**.