Question

c) after the first marble is drawn, for the second draw, how many marbles are left in the jar? what is the probability of drawing: d) a red marble on the second draw, given that a red marble was drawn first? hint: be careful with the denominator. how many marbles are in the jar? e) a blue marble on the second draw, given that a red marble was drawn first? for the remaining parts of this problem, instead suppose that the first marble is replaced (returned to the jar) before the second marble is drawn. f) after the first marble is drawn, for the second draw, how many marbles are left in the jar? what is the probability of drawing: g) a red marble on the second draw, given that a red marble was drawn first? h) a blue marble on the second draw, given that a red marble was drawn first?

Explanation:

Step1: Analyze non - replacement case for e

When a red marble is drawn first in non - replacement, there are 55 marbles left for the second draw. Let's assume there were originally $r$ red marbles and $b$ blue marbles with total marbles $n$. Since a red marble is drawn first, for the second draw to get a blue marble, the probability is the number of blue marbles left divided by the total number of marbles left. If we assume originally there were 24 red and 32 blue marbles ($n = 56$), after drawing a red marble first, the number of blue marbles is 32 and total marbles is 55. So the probability of drawing a blue marble on the second draw given a red marble was drawn first is $\frac{32}{55}$.

Step2: Analyze replacement case for g

When the first marble is replaced, the total number of marbles remains the same for the second draw as it was initially. Let's assume there are $r$ red marbles and total $n$ marbles. If we assume there are 24 red marbles and 56 total marbles, the probability of drawing a red marble on the second draw given a red marble was drawn first (since replacement occurs) is $\frac{24}{56}=\frac{3}{7}
eq\frac{22}{55}$.

Step3: Analyze replacement case for h

When the first marble is replaced, the total number of marbles is 56 for the second draw. Let's assume there are $b$ blue marbles and total $n$ marbles. If we assume there are 32 blue marbles and 56 total marbles, the probability of drawing a blue marble on the second draw given a red marble was drawn first (because of replacement) is $\frac{32}{56}=\frac{4}{7}
eq\frac{30}{55}$.

Answer:

e) $\frac{32}{55}$ is correct for non - replacement.
g) Incorrect, correct value is $\frac{\text{number of red marbles}}{\text{total number of marbles}}$ (e.g., if 24 red and 56 total, $\frac{24}{56}$).
h) Incorrect, correct value is $\frac{\text{number of blue marbles}}{\text{total number of marbles}}$ (e.g., if 32 blue and 56 total, $\frac{32}{56}$).