Question

please type in fractions like 2/3 (you do not need to reduce). a jar contains 56 marbles: 24 red marbles and 32 blue marbles. two marbles will be drawn from the jar. what is the probability of drawing: a) a red marble on the first draw? $\frac{24}{56}$ b) a blue marble on the first draw? $\frac{32}{56}$ the first marble is not placed back in the jar before the second marble is drawn. c) after the first marble is drawn, for the second draw, how many marbles are left in the jar? 55 what is the probability of drawing: d) a red marble on the second draw, given that a red marble was drawn first? $\frac{23}{55}$ 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? $\frac{31}{55}$

Explanation:

Step1: Analyze the situation

After drawing a red - marble first, there are 55 marbles left in the jar. The number of blue marbles remains 32 since only a red marble was removed.

Step2: Calculate the probability

The probability of drawing a blue marble on the second draw given that a red marble was drawn first is the number of blue marbles left divided by the total number of marbles left. So the probability is $\frac{32}{55}$.

Answer:

$\frac{32}{55}$