Question

There are two boxes containing only yellow and white pens.Box A has 7 white pens and 13 yellow pens.Box B has 6 white pens and 9 yellow pens.A pen is randomly chosen from each box.List these events from least likely to most likely.Event 1: choosing a white pen from Box A.Event 2: choosing a white pen from Box B.Event 3: choosing a white or yellow pen from Box B.Event 4: choosing a red pen from Box A.Least likelyMost likelyEvent | Event| Event EventEvent

Answer 1

Given:

Box A has 7 white pens and 13 yellow pens.

$\text{The sample space =7+13=20 pens}$$n(S_A)\text{ for A=20}$

Box B has 6 white pens and 9 yellow pens.

$\text{The sample space =6+9=15 pens}$

$n(S_B)\text{ for B=}15$

Event 1: choosing a white pen from Box A.

There are 7 white pens in Box A.

$n(1)=7$

The probability of getting a white pen from Box A is P(1).

$P(1)=\frac{n(1)}{n(S_A)}$$\text{ Substitute n(1)=7 and }n(S_A)=20,\text{ we get}$

$P(1)=\frac{7}{20}$

The probability of event 1 is 7/20

Event 2: choosing a white pen from Box B.

There 6 white pens in Box B.

$n(2)=6$

The probability of getting a white pen from Box B is P(2).

$P(2)=\frac{n(2)}{n(S_B)}$

$\text{ Substitute n(2)=6 and }n(S_B)=15,\text{ we get}$

$P(2)=\frac{6}{15}=\frac{2}{5}$

The probability of event 2 is 2/5.

Event 3: Choose a white or yellow pen from Box B.

There are 6 white pens and 9 yellow pens.

$n(3)=7+9=15$

The probability of getting a white pen or yellow pen from Box B is P(3).

$P(3)=\frac{n(3)}{n(S_B)}$

$\text{ Substitute n(3)=15 and }n(S_B)=15,\text{ we get}$

$P(3)=\frac{15}{15}=1$

The probability of event 3 is 1.

Event 4: Choose a red pen from Box A.

There are 0 red pens in Box A.

$n(4)=0$

The probability of getting a red pen from Box A is P(4).

$P(4)=\frac{n(4)}{n(S_A)}$

$\text{ Substitute n(4)=0 and }n(S_A)=20,\text{ we get}$

$P(4)=\frac{0}{20}=0$

The probability of event 4 is 0.

Recall the least likely means the slightest probability of occurring in relation to other events and the most likely means the higher the probability of occurring in relation to other events.

Event 4 is 0, so 0 probability it should take the place 1.

Event 3 is 1, so the most possible, it should take the place 4.

We need to compare Event 2 and Event 3.

7/20 and 2/5

Making the denominator 20, we get

$\frac{7}{20}\text{ and }\frac{2\times4}{5\times4}=\frac{8}{20}$$\frac{7}{20}<\frac{8}{20}$

So the order of the event from least to greatest is

$0<\frac{7}{20}<\frac{8}{20}<1$$\text{Event 4,Event 1,Event 2, Event 3}$