Problem 2.191 a researcher receives 93 containers of oxygen. of those containers, 20 have oxygen that is not ionized and the rest are ionized. two samples are randomly selected, without replacement, from the lot. round your answers to three decimal places (e. g. 98.765). (a) what is the probability that the first one selected is not ionized? (b) what is the probability that the second one selected is not ionized given that the first one was ionized? (c) how does the answer in part (b) change if samples selected were replaced prior to the next selection? find the probability. (d) what is the probability that both are ionized?

Step-by-step explanation:

Given that there are 20 non ionized containers and 73 ionized containers

Two samples are drawn without replacement

a) the probability that the first one selected is not ionized=

b)  the probability that the second one selected is not ionized given that the first one was ionized

= When first one was ionized we got left over as 20 and 72

Hence =

c) If with replacement left over 20 and 73 and hence prob = 0.215 as in part a

d) the probability that both are ionized=

the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195,   210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375,   390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, and 735.

the factors of 15 are 1, 3, 5, and 15.

hope that , fam!

-une, your fren : )

hi there!

question 1 answer: p/q can be terminating when q = 5^m * 2^n

question 2 answer: -1 and -1

question 3 answer: -4

question 4 answer: 5/12

hope this : )