Question

A box of 10 cellphones contains two yellow cellphones and eight green cellphones. Complete parts​ (a) through​ (d) below. Part 1 A. If two cellphones are randomly selected from the box without​ replacement, what is the probability that both cellphones selected will be​ green? ​P (both ​green) = 0.6222 or 0.6222 ​(Round to four decimal places as​ needed.) Part 2 B. If two cellphones are randomly selected from the box without​ replacement, what is the probability there will be one green cellphone and one yellow cellphone​ selected? ​P (1 green and 1 ​yellow) = ​(Round to four decimal places as​ needed.) C. If three cellphones are selected with replacement​ (the first cellphone is returned to the box after it is​ selected), what is the probability that all three will be​ yellow? P (three yellow) = ​(Round to four decimal places as​ needed.) D. If you were sampling with replacement​ (the first cellphone is returned to the box after it is​ selected), what would be the answers to​ (a) and​ (b)? P (both ​green) = ​(Round to four decimal places as​ needed.) P (green 1 and yellow 1) = ​(Round to four decimal places as​ needed.)

Answer 1

Let's calculate the probabilities step by step:

A. Probability that both cellphones selected will be green (without replacement):

P(both green) = (8/10) * (7/9)

P(both green) = 56/90

P(both green) = 28/45 ≈ 0.6222 (rounded to four decimal places)

So, the probability that both cellphones selected will be green is approximately 0.6222.

B. Probability that there will be one green cellphone and one yellow cellphone selected (without replacement):

P(1 green and 1 yellow) = (8/10) * (2/9) + (2/10) * (8/9)

P(1 green and 1 yellow) = (16/90) + (16/90)

P(1 green and 1 yellow) = 32/90 ≈ 0.3556 (rounded to four decimal places)

So, the probability of selecting one green and one yellow cellphone is approximately 0.3556.

C. Probability that all three cellphones will be yellow (with replacement):

P(three yellow) = (2/10) * (2/10) * (2/10)

P(three yellow) = 8/1000

P(three yellow) = 2/250

P(three yellow) = 1/125 ≈ 0.0080 (rounded to four decimal places)

So, the probability that all three cellphones will be yellow with replacement is approximately 0.0080.

D. If you were sampling with replacement:

For part (a), the probability of both cellphones being green remains the same as calculated in part (A), which is approximately 0.6222.

For part (b), the probability of selecting one green and one yellow cellphone with replacement is also the same as calculated in part (B), which is approximately 0.3556.