Question

DATA ANALYSES AND STATISTICSUnderstanding likelihoodThere are two boxes containing only blue and purple pens.Box A has 6 purple pens and 2 blue pens.Box B has 12 purple pens and 8 blue pens,A pen is randomly chosen from each box,List these events from least likely to most likelyEvent 1: choosing a purple or blue pen from Box B.Event 2: choosing a blue pen from Box B.Event 3: choosing a red pen from Box A.Event 4: choosing a purple pen from Box A.Least likelyMost likelyEvent ). Event | Event | .Event

Answer 1

Event 3, Event 2, Event 4, Event 1.

Step-by-step explanation

Given data:

Box A has 6 purple pens and 2 blue pens.

Box B has 12 purple pens and 8 blue pens.

Let X represents purple, Y represents blue, and Z represents red.

Box A: n(X) = 6, n(Y) = 2, and n(Z) = 0. Hence, n(S) = 6 + 2 + 0 = 8

Box B: n(X) = 12, n(Y) = 8, and n(Z) = 0. Hence, n(S) = 12 + 8 + 0 = 20

So if a pen is randomly chosen from each box, the probability of:

Event 1: choosing a purple or blue pen from Box B. will be:

`\begin{gathered} p(X\text{ or Y})=p(X)+p(Y)=(n(X))/(n(S))+(n(Y))/(n(S))=(12)/(20)+(8)/(20)=(12+8)/(20)=(20)/(20)=1 \\ \end{gathered}`

Event 2: choosing a blue pen from Box B will be:

`p(Y)=(8)/(20)=(1)/(5)=0.4`

Event 3: choosing a red pen from Box A will be:

`p(Z)=(n(Z))/(n(S))=(0)/(8)=0`

Event 4: choosing a purple pen from Box A wil be:

`p(X)=(n(X))/(n(S))=(6)/(8)=(3)/(4)=0.75`

If a pen is randomly chosen from each box, the list of the events from least likely to most likely are:

Event 3, Event 2, Event 4, Event 1.