Mathematics: In a list of households, own homes and do not own homes. Four households are randomly selected from these households. Find the probability that the number of households in these who own homes is exactly .

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In a list of households, own homes and do not own

silasjob09

07.02.2022

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

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

$A = ^{k} C_ 3$

Here C stands for combination hence we would be making use of the combination feature in our calculator

=> $A = ^{9} C_ 3$

=> A =84

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

$B = ^{u}C_1$

=> $B = ^{6}C_1$

=> B = 6

Generally the number of ways of selecting 4 household from the total number of 15 households on the list is mathematically represented as

$D = ^{n}C_4$

=> $D = ^{15}C_4$

=> D = 1365

Generally the probability that the number of households in the sample which own homes is exactly 3 is mathematically represented as

$P( X = 3) = \frac{A * B}{D}$

=> $P( X = 3) = \frac{85 * 6}{1365}$

X is here represent the number of household in the selected sample that own homes

=> P( X = 3) = 0.3736

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