The correct answer is option (E):
92%
Let's break down this problem using the principle of inclusion-exclusion and some careful calculations to determine the correct percentage.
First, let's denote the sets:
* T = Houses with TV
* A = Houses with AC
* W = Houses with Washing Machine
We're given the following information:
* Total houses surveyed = 1500
* |T| = 862
* |A| = 783
* |W| = 736
* Houses with only TV = 95
* Houses with only AC = 136
* Houses with only Washing Machine = 88
* Houses with all three (T ∩ A ∩ W) = 398
We want to find the percentage of houses with all three equipments to the houses with only two equipments.
First, let's find the houses with only two equipments. We know the number of houses with all three. To find houses with only two equipments we subtract the houses having all three equipments from the intersections of any two of the equipment.
We know:
* |T ∩ A ∩ W| = 398
* Only T = 95
* Only A = 136
* Only W = 88
To get the number of houses having only T and A, subtract the number of houses having all three equipment (398), from all houses with TV and AC. We don't have the specific numbers for the intersection of pairs directly, but we can utilize the principle of inclusion-exclusion. But first, we need to find the houses that have only two items.
Let's find the total houses with TV. We know it includes only TV, TV with only AC, TV with only Washing Machine and TV with AC and Washing Machine(all three)
Houses having only TV = 95
Houses having all three (T ∩ A ∩ W) = 398
So, TV and AC (only) = x
TV and Washing Machine (only) = y
So, |T| = Only TV + (only TA) + (only TW) + (TAW)
862 = 95 + x + y + 398
862 = 493 + x + y
x + y = 369
Similarly:
Houses having only AC = 136
AC and TV (only) = x
AC and Washing Machine (only) = z
Houses having all three (T ∩ A ∩ W) = 398
So, |A| = Only AC + (only TA) + (only AW) + (TAW)
783 = 136 + x + z + 398
783 = 534 + x + z
x + z = 249
Similarly:
Houses having only Washing Machine = 88
TV and Washing Machine (only) = y
AC and Washing Machine (only) = z
Houses having all three (T ∩ A ∩ W) = 398
So, |W| = Only W + (only TW) + (only AW) + (TAW)
736 = 88 + y + z + 398
736 = 486 + y + z
y + z = 250
Solving for the values of x, y, and z. We have:
x + y = 369
x + z = 249
y + z = 250
Adding all the equations :
2x + 2y + 2z = 868
x + y + z = 434
Now:
z = 434 - (x + y) = 434 - 369 = 65
y = 434 - (x + z) = 434 - 249 = 185
x = 434 - (y + z) = 434 - 250 = 184
Hence:
Houses having only TA = x = 184
Houses having only TW = y = 185
Houses having only AW = z = 65
So, total houses having only two equipments = 184 + 185 + 65 = 434
Houses having all three equipments = 398
Percentage = (Houses with all three equipments / Houses with only two equipments) * 100
= (398 / 434) * 100
= 91.69% which is approximately 92%
Therefore, the correct answer is 92%.