Question

Suppose you have a box with 10 red balls, 4 green balls, and 6 blue balls. Answer the following questions. Think about the questions below. They look similar, but how are they different?

a) You randomly select two balls from the box. What is the probability of selecting two red balls?

b) Suppose you are asked to draw a ball, return whatever you picked, and draw another ball. This is called sampling with replacement. What is the probability of selecting two red balls?

c) Suppose you are asked to select two balls in sequence, without returning any back to the box. This is called sampling without replacement. What is the probability of selecting two red balls?

d) Suppose you selected two balls in sequence, without returning either to the box. Given that you selected a red ball in the first draw, what is the probability of getting a red ball again in the second draw?

Answer #1

a) You randomly select two balls from the box. What is the probability of selecting two red balls?

P(Red=2) = 10C2 / 20C2 =10*9/20*19 = **9/38** =
**0.2368**

b) P(Selecting 2 red balls with replacement) = P(1 Red ball out
of 20) * P(1 Red ball out of 20) = 10/20 * 10/20
=**1/4**=**0.25**

c)P(Selecting 2 red balls without replacement) = 10C2 / 20C2
=10*9/20*19 = **9/38** = **0.2368**

d)you selected two balls in sequence, without returning either
to the box. Given that you selected a red ball in the first draw
the probability of getting a red ball again in 2nd draw= Red balls
remaining/total balls remaining = **9/19=0.4737**

*Hope the above answer has helped you in understanding the
problem. Please upvote the ans if it has really
helped you. Good Luck!!*

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