Question

Suppose you have a bag with ten balls in it, and each ball has a digit printed on it:

0,1,2,3,4,5,6,7,8,9. So there is one ball for each of the ten digits 0-9.

a) If you pull out three balls and set them aside, what is the probability that you pull out the

numbers “5”, “0”, and “4”, in that order? (“5” is your first pull, “0” is your second, and “4” is

your third).

b) For part (a), what is the probability that your three numbers consist of “5”, “0”, and “4” in

any order? Express your answer as a fully reduced fraction.

c) Suppose instead you take a ball out, record its value, then return it into the bag, then repeat

this process three total times. What is the probability that you will select the 5, the 0, and the 4,

in that order?

d) In part (c), what is the probability that your three selections consist of the number 4 twice,

and the number 8 once, in any order?

Answer 1

PLEASE FIND THE ATTACHMENT BELOW

(1) 3 To take out the probablity we will flot draw one ball that should be s and aside then we drow second ball that should o and bett aside and then draw third ball that should be u = 3100, Vio) (3/10) 3 x 9c. US)" (4 8 x 26, 2) (2/2)* > 3.059295096 (6) Favougable outcome. a) 504 54o, e os 4,40 4.5 ust €405 -) 6.059295086 x6 >.0.355770576. C To take out hrobablety we draw one second its value and fout is back to the bage. so the Probability it will 5,0 4 in that as dis- => 106, Vio)" (200) X10C, CYD) XEV0J X104,) *09:25 3. 0:058149737 Favourdelele out come. = 448 4 84 8414 = 0.058149737 x 3 = 0.174449211 .