Question

Colorful disksA jar contains $36$disks: $9$each of four colors—red, green, blue, and Page Number: $328$yellow. Each set of disks of the same color is numbered from $1$to $9$. Suppose you draw one disk at random from the jar. Define events $R$: get a red disk, and $N$: get a disk with the number $9$.

a. Make a two-way table that describes the sample space in terms of events $R$and $N$.

b. Find $P\left(R\right)$and $P\left(N\right)$.

c. Describe the event “$R$and $N$” in words. Then find the probability of this event.

d. Explain why $P(RorN)\ne P\left(R\right)+P\left(N\right)$ Then use the general addition rule to compute$P(RorN).$

Short answer

a. A two − way table describing sample space in terms of events $R$ and $N$.

	Nine	Not Nine	Total
Red	$1$	$8$	$9$
Not red	$3$	$24$	$27$
Total	$4$	$32$	$36$

b. The probabilities of both events $R$and $N$is$0.1111$

c. The probability of "$RandN$" is$0.0278$

d. The probability of "$RorN$" is$0.3333$

Step-by-step solution

Part (a) Step 1 : Given Information

We have to determine two − way table describing sample space in terms of events $R$ and $N$.

Part (a) Step 2 : Simplification

$R$: grab yourself a red disc

Get a disc with the number $9$on it $N$.

In the case of a two-way table,
It is necessary for us to create two rows and two columns.
Let's call the rows and columns after the $R$and $N$events.

Rows : red , not red

Columns : nine , not nine

We must write $36$in the bottom left corner of the table because there are $36$discs in total.

We are aware of this.
There are $9$red, blue, green, and yellow discs in total.

This denotes

The discs are crimson in hue and there are nine of them and the remaining $27$blue, green, and yellow discs.

Wealsounderstandthat

Each colour disc is assigned a number from $1$to $9$.

This denotes

There is only one disc, which is crimson in colour and has the number nine on it.

And

The last three discs are numbered 9 and come in a variety of colours.

This suggests that

Only one disc is numbered $9$and is tinted red; the remaining eight discs are numbered $1-8$.

Due to the fact that $3$discs of different colours are numbered 9, the remaining $24$discs of different colours are numbered $1-8$.

Two − way table is shown as:

	Nine	Not Nine	Total
Red	$1$	$8$	$9$
Not red	$3$	$24$	$27$
Total	$4$	$32$	$36$

Part (b) Step 1 : Given Information

We have to determine the probabilities of both events $R$ and $N$.

Part (b) Step 2 : Simplification

R:grabyourselfareddisk

Get a disk with the number $9$on it ($N$).

We are aware of this.
The colour red appears on $9$of the $36$discs.

The probability is calculated by dividing the number of favourable outcomes by the total number of possible possibilities.

$P\left(R\right)=\frac{Number\hspace{0.17em}of\hspace{0.17em}favourable\hspace{0.17em}outcomes}{Number\hspace{0.17em}of\hspace{0.17em}possible\hspace{0.17em}outcomes}=\frac{9}{36}=\frac{1}{4}=0.25$

Wealsounderstandthat

Each colour disc is assigned a number from $1$to $9$.

$4$discs with $4$colours are numbered $9$out of $36$discs.

$P\left(N\right)=\frac{Number\hspace{0.17em}of\hspace{0.17em}favourable\hspace{0.17em}outcomes}{Number\hspace{0.17em}of\hspace{0.17em}possible\hspace{0.17em}outcomes}=\frac{4}{36}=\frac{1}{9}\approx 0.1111$

Part (c) Step 1 : Given Information

We have to discuss the event “$R$ and $N$” and also find its probability.

Part (c) Step 2 : Simplification

$R$: grab yourself a red disc

Get a disc with the number $9$on it ($N$).

The letters "$R$and $N$" stand for the fact that the disc is numbered $9$and is red in hue.

We are aware of this.

There are $9$red discs in all, out of a total of $36$discs.

Since each colour disc is numbered $1$to $9$, there are a total of $9$colour discs.

This denotes

Among the $36$discs, there is only one that is numbered $9$and red in colour and has the event "$R$and $N$."

Now,

In terms of the likelihood,

The number of favourable results is divided by the total number of options.

$P(R\hspace{0.17em}and\hspace{0.17em}N)=\frac{Number\hspace{0.17em}of\hspace{0.17em}favourable\hspace{0.17em}outcomes}{Number\hspace{0.17em}of\hspace{0.17em}possible\hspace{0.17em}outcomes}\approx 0.0278$

Part (d) Step 1 : Given Information

We have to discuss the probability of event “$RorN$” is not equal to the sum of individual probabilities of both events. Furthermore, general addition rule to be used for the probability of event “$RorN$”.

Part (d) Step 2 : Simplification

$R$: grab yourself a red disc

Get a disc with the number $9$on it ($N$).

If there are two events,

Rule of thumb for addition:

$P(A\hspace{0.17em}or\hspace{0.17em}B)=P\left(A\right)+P\left(B\right)-P(A\hspace{0.17em}and\hspace{0.17em}B)$

According to the declaration,

$P(R\hspace{0.17em}or\hspace{0.17em}N)\ne P\left(R\right)+P\left(N\right)$

ThereasonforthisisthattheoccurrencesRandNarenotmutuallyexclusive.

That is to say,

It is feasible for both occurrences to occur at the same moment.

as a result

$P(R\hspace{0.17em}and\hspace{0.17em}N)\ne 0$

Parts ($b$) and ($c$),

We've got

Probabilities,

For the $R$event: $P\left(R\right)=\frac{9}{36}$

For event N: $\hspace{1em}\hspace{1em}P\left(N\right)=\frac{4}{36}$

For event “$RandN$” :- $P(R\hspace{0.17em}and\hspace{0.17em}N)=\frac{1}{36}$

For any two events, use the following generic addition rule:

$$\begin{gathered} P(R\hspace{0.17em}or\hspace{0.17em}N)=P\left(R\right)+P\left(N\right)-P(R\hspace{0.17em}and\hspace{0.17em}N) \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{9}{36}+\frac{4}{36}-\frac{1}{36}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{12}{36}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\approx 0.3333 \end{gathered}$$