Question

AllElectronics carries 1000 products, P1, . . . , P1000. Consider customers Ada, Bob, and Cathy such that Ada and Bob purchase three products in common, P1,P2, and P3. For the other 997 products, Ada and Bob independently purchase seven of them randomly. Cathy purchases 10 products, randomly selected from the 1000 products. In Euclidean distance, what is the probability that dist(Ada,Bob) > dist(Ada,Cathy)

Answer 1

Answer:

The probability that dist(Ada,Bob)>(Ada,Cathy)  is very small as there is very large number of range to choose the product ==4.7*10^-9.

Step-by-step explanation:

Given:

Ada ,bob and cathy purchase electronics carries

Ada and bob commonly take 3 products and 7 independently.

And Cathy take 10  products on its own .

To Find:

probability that    dist(Ada,bob)>dis(Ada,Cathy)?

Solution:

Using Euclidean distance  is distance formula used in coordinate geometry simply known as Distance formula,

this problem is related to Euclidean Distance and Jaccard Similarity in Data mining.

1st calculate probability for x such that ,

$3\leq x\leq 10$   as there are 3 common products.

P($3\leq x\leq 10$)

=$\frac{7C(x-3)*990C(10-x)}{997C7}$.............. where x=3,4,5....10. ..........(equation 1).

Now calculate for each term,we get

When

x=3,P(x=3)=0.95

x=4,P(x=4)=$6.8*10^{-3}$

x=5,P(x=5)=$4.1*10^{-5}$

x=6,P(x=6)=2.1*10^-7

x=7,P(x=7)=8.5*10^-10

x=8,P(x=8)=2.6*10^-12

x=9,P(x=9)=5.2*10^-15

x=10,P(x=10)=5.3*10^-18.

Now calculating the Euclidean distance,

It is distance between two points ,

So there are total of 2 points as Ada and bob

they have 3 products in common

and 7 independent products ,7  Ada and 7 bob

Total of 17 products .

1,2,3,4,5,6..........,16,17.

Consider each product number as distance between them ,

(Suppose 5 product and 1 product distance will be 4)

Similarly,

Suppose Ada is at 3rd number at the  3 product (as they have 3 product same.)

and bob  at product 17.

Hence when 3 products are similar distance between Ada and bob will be of 14.

Euclidean distance =$\sqrt{14}$.

Hence the Jaccard similarity =(Ada intersection Bob)/(Ada union bob)

=3/14

When 4 products are same means both will selected 6 and 6 independent product so that  the each one will get 10 products i.e. starting condition should remain same .

Hence now  bob will be at 16th term as it will take one more same product in between them

So no of same products will be 4,

Hence Ada will be at 4th term and bob will be at 16

So Euclidean distance =$\sqrt{12}$.

Similar For Next terms we can conclude as follows:

When

X=5 , dist(ada,bob)=$\sqrt{10}$,

X=6,dist(Ada,Bob)=$\sqrt{8}$

X=7,dist(Ada,Bob)=$\sqrt{6}$

X=8,dist(Ada,Bob)=$\sqrt{4}$

X=9,dist(Ada,Bob)=$\sqrt{2}$

X=10,dist(Ada,Bob)=$\sqrt{0}$.

Now for( Ada and cathy)

Here X ranges different but use same concept as above

Each term analog to the distance between them

Suppose 1st and 3rd term distance will be 2

First calculate

P($1\leq x\leq 10$) as Cathy selects 10 products with no common between them.

P($1\leq x\leq 10$)

=$\frac{10Cx*990C(10-x)}{1000C10}$..................equation (2)

Calculate for each term As x=1,2,3...8,9,10.

Hence

P(X=1)=9.23*10^-3  P(X=5)=3*10^-11     P(X=9)=3.8*10^-21

P(X=2)=8.4*10^-5   P(X=6)=1.5*10^-13   P(X=10)=3.8*10^-21

P(X=3)=6.9*10^-7   P(X=7)=6.1*10^-16

P(X=4)=4.9*10^-9   P(X=8)=1.9*10^-18

So Ada will have 10 products and Cathy will have 10 products

Namely,

1,2,3,4,5.......18,19,20.

So suppose 1 product is same between them will be ,

both will have 1 product so remaining will be 19 products.

Jaccard similarity =1/19

Distance to reach 1 to 19th product will be 18

So Euclidean distance =$\sqrt{18}$

For next when they will 2 products in same remaining will be 18

Jaccard similarity =2/18

And Distance to reach  2 to 18 th product will be  16

Euclidean distance =$\sqrt{16}$

Similar for  other

When

x=3 dist(Ada, Cathy)=$\sqrt{14}$

x=4 dist(Ada, Cathy)=$\sqrt{12}$

x=5  dist(Ada, Cathy)=$\sqrt{10}$

x=6  dist(Ada, Cathy)=$\sqrt{8}$

x=7  dist(Ada, Cathy)=$\sqrt{6}$

x=8  dist(Ada, Cathy)=$\sqrt{4}$

x=9  dist(Ada, Cathy)=$\sqrt{2}$

x=10  dist(Ada, Cathy)=$\sqrt{0}$

This sqrt(0) means both are holding same products hence they are at same point on the graph so distance with itself will be zero.

Now the Probability of distance of dist(Ada,Bob)>dist(Ada,cathy) will be

=multiplying both the probabilities equations (Adding each term probabilities and multiplying )

=Equation(1) *Equation( 2).

=Summation Of P(3≤x≤10)*summation of P(1≤x≤10)

=4.7*10^-9.

In larger number of product event of in large space ,it is difficult( less likely)  that they will chose same product .