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 ,
as there are 3 common products.
P()
=
.............. 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)=
x=5,P(x=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 =
.
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 =
.
Similar For Next terms we can conclude as follows:
When
X=5 , dist(ada,bob)=
,
X=6,dist(Ada,Bob)=
X=7,dist(Ada,Bob)=
X=8,dist(Ada,Bob)=
X=9,dist(Ada,Bob)=
X=10,dist(Ada,Bob)=
.
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(
) as Cathy selects 10 products with no common between them.
P()
=
..................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 =
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 =
Similar for other
When
x=3 dist(Ada, Cathy)=
x=4 dist(Ada, Cathy)=
x=5 dist(Ada, Cathy)=
x=6 dist(Ada, Cathy)=
x=7 dist(Ada, Cathy)=
x=8 dist(Ada, Cathy)=
x=9 dist(Ada, Cathy)=
x=10 dist(Ada, Cathy)=
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 .
