Complete Question
In a list of 15 households, 9 own homes and 6 do not own homes. Four households
are randomly selected from these 15 households. Find the probability that the
number of households in the sample which own homes is exactly 3
Answer:
Step-by-step explanation:
From the question we are told that
The number of household is n = 15
The number of household that own homes is k = 9
The number of household that do not own homes is u = 6
The number of households randomly selected is q = 4
Generally the number of ways of selecting 3 household from the total number of 9 households that own homes is mathematically represented as
[tex]A = ^{k} C_ 3[/tex]
Here C stands for combination hence we would be making use of the combination feature in our calculator
=> [tex]A = ^{9} C_ 3[/tex]
=> [tex]A =84[/tex]
Generally the number of ways of selecting 1 household from the total number of 6 households that do not own homes is mathematically represented as
[tex]B = ^{u}C_1[/tex]
=> [tex]B = ^{6}C_1[/tex]
=> [tex]B = 6[/tex]
Generally the number of ways of selecting 4 household from the total number of 15 households on the list is mathematically represented as
[tex]D = ^{n}C_4[/tex]
=> [tex]D = ^{15}C_4[/tex]
=> [tex]D = 1365[/tex]
Generally the probability that the number of households in the sample which own homes is exactly 3 is mathematically represented as
[tex]P( X = 3) = \frac{A * B}{D}[/tex]
=> [tex]P( X = 3) = \frac{85 * 6}{1365}[/tex]
X is here represent the number of household in the selected sample that own homes
=> [tex]P( X = 3) = 0.3736[/tex]
