Independent solution

How to solve this Discrete Random Variables question

Setup

Setup

Let X and Y be the two technicians' counts. They are independent and share the stated probability mass function.

Pr⁡(X=k)=Pr⁡(Y=k)=(1,2,4,8,3,2)k20,k=0,1,…,5\Pr(X=k)=\Pr(Y=k)=\frac{(1,2,4,8,3,2)_k}{20},\qquad k=0,1,\ldots,5

Model

Model

A total of at least eight is possible only for six ordered pairs. Independence turns each joint probability into a product of marginal masses.

{(x,y):x+y≥8}={(3,5),(4,4),(5,3),(4,5),(5,4),(5,5)}\{(x,y):x+y\ge 8\}=\{(3,5),(4,4),(5,3),(4,5),(5,4),(5,5)\}
Pr⁡(X=x,Y=y)=Pr⁡(X=x)Pr⁡(Y=y)\Pr(X=x,Y=y)=\Pr(X=x)\Pr(Y=y)

Compute

Compute

Use the distribution weights directly and retain both orders whenever the counts differ.

Pr⁡(X+Y≥8)=8⋅2+3⋅3+2⋅8+3⋅2+2⋅3+2⋅2202\Pr(X+Y\ge 8)=\frac{8\cdot2+3\cdot3+2\cdot8+3\cdot2+2\cdot3+2\cdot2}{20^2}
Pr⁡(X+Y≥8)=57400=0.1425\Pr(X+Y\ge 8)=\frac{57}{400}=0.1425

Answer

Answer

The required probability is 57/400.

57400(E)\boxed{\frac{57}{400}\quad\text{(E)}}