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 part e

In non - replacement, after drawing a red marble first, there are 55 marbles left in the jar. Let the number of blue marbles be $b$. The probability of drawing a blue marble on the second draw given a red marble was drawn first is $\frac{b}{55}$.

Step2: Analyze replacement case for part g

When the first marble is replaced, the total number of marbles in the jar remains the same for the second draw. Let the number of red marbles be $r$ and total marbles be $n$. The probability of drawing a red marble on the second draw given a red marble was drawn first is $\frac{r}{n}$. Since the first marble is replaced, the probability is the same as the probability of drawing a red marble on the first draw.

Step3: Analyze replacement case for part h

When the first marble is replaced, the total number of marbles in the jar is $n$. Let the number of blue marbles be $b$. The probability of drawing a blue marble on the second draw given a red marble was drawn first is $\frac{b}{n}$.

Answer:

e) Let's assume the number of blue marbles is 31 (since the denominator is 55 and we are calculating the probability of drawing a blue marble after a red one in non - replacement). The correct probability is $\frac{31}{55}$.
g) If the first marble is replaced, assume there are 22 red marbles and 56 total marbles. The probability of drawing a red marble on the second draw given a red marble was drawn first is $\frac{22}{56}=\frac{11}{28}$.
h) If the first marble is replaced, assume there are 30 blue marbles and 56 total marbles. The probability of drawing a blue marble on the second draw given a red marble was drawn first is $\frac{30}{56}=\frac{15}{28}$.