To find the probability of drawing two red marbles in succession without replacement from a bag containing 11 red marbles and 4 green marbles (15 marbles total), we proceed step by step:

Step 1: Probability of drawing a red marble first

• There are 11 red marbles out of 15 total marbles.
• So, the probability of drawing a red marble first is: $P(\text{Red 1st}) = \frac{11}{15}$

Step 2: Probability of drawing a red marble second without replacement

• After drawing one red marble, there are now 10 red marbles left, and only 14 total marbles remain.
• So, the probability of drawing a second red marble is: $P(\text{Red 2nd | Red 1st}) = \frac{10}{14}$

Step 3: Multiply the probabilities

The overall probability of drawing two red marbles in a row without replacement is: $P(\text{Red 1st and Red 2nd}) = \frac{11}{15} \times \frac{10}{14}$ $P(\text{Red 1st and Red 2nd}) = \frac{110}{210} = \frac{11}{21}$

Thus, the probability of drawing two red marbles in succession without replacement is $\frac{11}{21}$.

Would you like further details or any additional questions?

Here are 5 related questions for further practice:

1. What is the probability of drawing two green marbles in a row without replacement?
2. What are the odds of drawing a red marble first and a green marble second without replacement?
3. If one marble is drawn and replaced, what are the odds of drawing two red marbles in a row?
4. How does the probability change if there are 6 green and 9 red marbles?
5. What is the probability of drawing one red and one green marble in any order without replacement?

Tip: When calculating probabilities without replacement, remember that the total number of items decreases after each draw.