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2 years ago

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An apartment building has 20 two-bedroom apartments and 8 three-bedroom apartments. If 4 apartments are chosen at random to be r

edecorated, what are the probabilities a) All will be two-bedroom apartments

2 answers:

seraphim [82]2 years ago

7 0

The answer is 4 of 28. i think so

ankoles [38]2 years ago

5 0

A) This is sampling without replacement.
$P(all\ 2\ bedroom)=\frac{20C4}{28C4}=\frac{20\times19\times18\times17}{28\times27\times26\times25}$

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Suppose approximately 75% of all marketing personnel are extroverts, whereas about 70% of all computer programmers are introvert

Setler [38]

Answer:

$P(x \ge 5) = 1.000$ ---- At least 5 from marketing departments are extroverts

$P(x=15) = 0.013$ ---- All from marketing departments are extroverts

$P(x = 0) = 0.002$ ---------- None from computer programmers are introverts

Step-by-step explanation:

See comment for complete question

The question is an illustration of binomial probability where

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$(a):\ P(x \ge 5)$

$n = 15$ --- marketing personnel

$p = 75\%$ --- proportion that are extroverts

Using the complement rule, we have:

$P(x \ge 5) = 1 - P(x < 5)$

So, we have:

$P(x < 5) =P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)$

$P(x = 0) = ^{15}C_{0} * (75\%)^0 * (1 - 75\%)^{15 - 0} = 1 * 1 * (0.25)^{15} = 9.31 * 10^{-10}$

$P(x = 1) = ^{15}C_{1} * (75\%)^1 * (1 - 75\%)^{15 - 1} = 15* (0.75)^1 * (0.25)^{14} = 4.19 * 10^{-8}$

$P(x = 2) = ^{15}C_{2} * (75\%)^2 * (1 - 75\%)^{15 - 2} = 105* (0.75)^2 * (0.25)^{13} = 8.80 * 10^{-7}$

$P(x = 3) = ^{15}C_{3} * (75\%)^3 * (1 - 75\%)^{15 - 3} = 455* (0.75)^2 * (0.25)^{12} = 0.0000153$

$P(x = 4) = ^{15}C_{4} * (75\%)^4 * (1 - 75\%)^{15 - 4} = 1365 * (0.75)^4 * (0.25)^{11} = 0.000103$

So, we have:

$P(x < 5) = (9.31 * 10^{-10}) + (4.19 * 10^{-8}) + (8.80 * 10^{-7}) + 0.0000153 + 0.000103$

$P(x < 5) = 0.00011922283$

Recall that:

$P(x \ge 5) = 1 - P(x < 5)$

$P(x \ge 5) = 1 - 0.00011922283$

$P(x \ge 5) = 0.9998$

$P(x \ge 5) = 1.000$ --- approximated

$(b)\ P(x = 15)$

$n = 15$ --- marketing personnel

$p = 75\%$ --- proportion that are extroverts

So, we have:

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$P(x=15) = ^{15}C_{15} * 0.75^{15} * (1 - 0.75)^{15-15}$

$P(x=15) = 1 * 0.75^{15} * (0.25)^{0$

$P(x=15) = 0.013$

$(c)\ P(x = 0)$

$n=5$ ---------- computer programmers

$p = 70\%$ --- proportion that are introverts

So, we have:

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$P(x = 0) = ^{5}C_0 * (70\%)^0 * (1 - 70\%)^{5-0}$

$P(x = 0) = 1 * 1 * (0.30)^5$

$P(x = 0) = 0.002$

3 0

2 years ago

Plz solve fast!<br> what is the answer of this quation.

Alex_Xolod [135]

49 is the answer to ur question

3 0

2 years ago

TSolve the system of equations below by graphing. What is the solution rounded to the nearest tenth? (–1.1, 3.2) (0.2, 2.8) (0.7

igor_vitrenko [27]

Answer: Answer:The solution to the given system of equations is (2.8,0.1)

Step-by-step explanation: I got it correct on the Unit Test

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2 years ago

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Can someone help me please?

MAVERICK [17]

Good. I hope you get locked, most mainly banned. Nobody needs you, “FrIsK”

4 0

2 years ago

Please answer both question fully.

BARSIC [14]

Step-by-step explanation:

<h3>Part A</h3>

<u>Angle measures:</u>

a = x = z = 51° (as vertical angles and corresponding angles)
b = d = y = w = 180° - 51° = 129° (as supplementary angles and corresponding angles)

<h3>Part B</h3>

Supplementary angles for a linear pair and sum to 180°.

<u>Examples are:</u>

a and b or
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7 0

1 year ago

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