To find the probability that both marbles drawn are red, we will use the rules of probability for dependent events (since the draws are without replacement).
Let's denote:
- [tex]\( R_1 \)[/tex] as the event of drawing the first red marble.
- [tex]\( R_2 \)[/tex] as the event of drawing the second red marble given that the first red marble has been drawn.
Here's a step-by-step breakdown of the solution:
1. Determine the total number of marbles:
The jar contains [tex]\( 14 \)[/tex] red marbles and [tex]\( 19 \)[/tex] blue marbles.
[tex]\[
\text{Total marbles} = 14 + 19 = 33
\][/tex]
2. Calculate the probability of drawing the first red marble ([tex]\( P(R_1) \)[/tex]):
The probability of drawing one red marble from the jar is:
[tex]\[
P(R_1) = \frac{\text{Number of red marbles}}{\text{Total number of marbles}} = \frac{14}{33}
\][/tex]
3. Determine the number of marbles remaining after drawing the first red marble:
If one red marble is drawn, then there are [tex]\( 33 - 1 = 32 \)[/tex] marbles left.
4. Calculate the probability of drawing the second red marble ([tex]\( P(R_2 \mid R_1) \)[/tex]):
Given that the first marble drawn is red, there are now [tex]\( 14 - 1 = 13 \)[/tex] red marbles left out of the remaining [tex]\( 32 \)[/tex] marbles.
[tex]\[
P(R_2 \mid R_1) = \frac{\text{Remaining red marbles}}{\text{Remaining total marbles}} = \frac{13}{32}
\][/tex]
5. Find the combined probability of both events occurring:
The probability of both marbles being red (drawing two red marbles in succession without replacement) is the product of the two probabilities:
[tex]\[
P(R_1 \text{ and } R_2) = P(R_1) \times P(R_2 \mid R_1)
\][/tex]
[tex]\[
P(R_1 \text{ and } R_2) = \frac{14}{33} \times \frac{13}{32}
\][/tex]
6. Perform the multiplication and simplify:
[tex]\[
P(R_1 \text{ and } R_2) = \frac{14 \times 13}{33 \times 32} = \frac{182}{1056}
\][/tex]
7. Simplify the fraction (if possible) and/or convert to a decimal:
Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor (GCD), which in this case is 2:
[tex]\[
\frac{182 \div 2}{1056 \div 2} = \frac{91}{528}
\][/tex]
In decimal form, this is approximately:
[tex]\[
\frac{91}{528} \approx 0.1723
\][/tex]
So, the probability that both marbles drawn are red is:
[tex]\[
\boxed{\frac{91}{528}} \text{ or } \boxed{0.1723}
\][/tex]
