Question

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 .

Answer 1

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  

        $A = ^{k} C_ 3$

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

=>      $P( X = 3) = 0.3736$