Question

A box of 8 cellphones contains two yellow cellphones and six green cellphones. Complete parts (a) through (d) below.
a. If two cellphones are randomly selected from the box without replacement, what is the probability that both cellphones selected will be green?
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?
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?
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)?

Answer 1

Probabilities are used to determine the chance of an event. The following are the summary of the solution.

• The probability that the two selected cellphones are green (without replacement) is 15/28

• The probability that one green and one yellow is selected (without replacement) is 3/7
• The probability that all three cellphones are yellow (with replacement) is 1/64
• The probability that the two cellphones are green (with replacement) is 9/16
• The probability that one green and one yellow is selected (with replacement) is 3/8

Given that:

$n = 8$

$G = 6$ --- Green

$Y = 2$ --- Yellow

(a) Probability that the two cellphones are green (without replacement).

Since the cellphone is not replaced, the probability is calculated as follows:

$Pr = \frac Gn \times \frac{G - 1}{n-1}$

So, we have:

$Pr = \frac 68 \times \frac{6 - 1}{8-1}$

$Pr = \frac 68 \times \frac 57$

$Pr = \frac{30}{56}$

$Pr = \frac{15}{28}$

Hence, the probability that the two cellphones are green (without replacement) is 15/28

(b) Probability that one green and one yellow is selected (without replacement).

Since the cellphone is not replaced, the probability is calculated as follows:

$Pr = \frac Gn \times \frac{Y}{n-1} + \frac Yn \times \frac{G}{n-1}$  ---- The subtraction means the cellphones are not replaced

This gives

$Pr = \frac 68 \times \frac{2}{8-1} + \frac 28 \times \frac{6}{8-1}$

$Pr = \frac 34 \times \frac{2}{7} + \frac 14 \times \frac{6}{7}$

$Pr = \frac 32 \times \frac17 + \frac 12 \times \frac 37$

$Pr = \frac{3}{14} + \frac{3}{14}$

Take LCM

$Pr = \frac{3+3}{14}$

$Pr = \frac{6}{14}$

$Pr = \frac{3}{7}$

Hence, the probability that one green and one yellow is selected (without replacement) is 3/7

(c) Probability that the all three cellphones are yellow (with replacement).

Since the cellphone is replaced, the probability is calculated as follows:

$Pr = \frac Yn \times \frac Yn \times \frac Yn$

So, we have:

$Pr = \frac 28 \times \frac 28 \times \frac 28$

$Pr = \frac 14 \times \frac 14 \times \frac 14$

$Pr = \frac 1{64}$

Hence, the probability that all three cellphones are yellow (with replacement) is 1/64

(d1) Probability that the two cellphones are green (with replacement).

Since the cellphone is replaced, the probability is calculated as follows:

$Pr = \frac Gn \times \frac{G}{n}$

So, we have:

$Pr = \frac 68 \times \frac{6}{8}$

$Pr = \frac 34 \times \frac{3}{4}$

$Pr = \frac{9}{16}$

Hence, the probability that the two cellphones are green (with replacement) is 9/16

(d2) Probability that one green and one yellow is selected (with replacement).

Since the cellphone is replaced, the probability is calculated as follows:

So, we have:

$Pr = \frac Gn \times \frac{Y}{n} + \frac Yn \times \frac{G}{n}$

This gives

$Pr = \frac 68 \times \frac{2}{8} + \frac 28 \times \frac{6}{8}$

$Pr = \frac 34 \times \frac{1}{4} + \frac 14 \times \frac{3}{4}$

$Pr = \frac 3{16} + \frac{3}{16}$

Take LCM

$Pr = \frac {3+3}{16}$

$Pr = \frac {6}{16}$

$Pr = \frac {3}{8}$

Hence, the probability that one green and one yellow is selected (with replacement) is 3/8

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