Question A wheel has 5 slices numbered 1 to 5, some grey and some white. Find $P(X)$ , the probability the wheel stops on a white slice, and $P(\text{not }X)$ , the probability it stops on a non-white slice.

Studdy Solution

STEP 1

Assumptions
1. The wheel has 5 equally sized slices numbered from 1 to 5.
2. The slices numbered 2, 4, and 5 are grey.
3. The slices numbered 1 and 3 are white.
4. The wheel is spun and stops on a slice at random.
5. Event $X$ is the wheel stopping on a white slice.
6. Event not $X$ is the wheel stopping on a slice that is not white.
7. The probability of the wheel stopping on any given slice is equal.

STEP 2

To determine the outcomes contained in event $X$ , identify the slices that are white.

STEP 3

Since slices 1 and 3 are white, these are the outcomes for event $X$ .

STEP 4

Fill in the outcomes for event $X$ in the table.
\begin{tabular}{|c|c|c|c|c|c|c|} \hline \multirow{2}{}{ Event } & \multicolumn{5}{|c|}{ Outcomes } & \multirow{2}{}{ Probability } \\ \cline { 2 - 6 } & 1 & 2 & 3 & 4 & 5 & \\ \hlineX & \checkmark & & \checkmark & & & P(X)=\square \\ \hline not X & & & & & & P(\operatorname{not} X)=\square \\ \hline \end{tabular}

STEP 5

To determine the outcomes contained in event not $X$ , identify the slices that are not white, which are grey.

STEP 6

Since slices 2, 4, and 5 are grey, these are the outcomes for event not $X$ .

STEP 7

Fill in the outcomes for event not $X$ in the table.
\begin{tabular}{|c|c|c|c|c|c|c|} \hline \multirow{2}{}{ Event } & \multicolumn{5}{|c|}{ Outcomes } & \multirow{2}{}{ Probability } \\ \cline { 2 - 6 } & 1 & 2 & 3 & 4 & 5 & \\ \hlineX & \checkmark & & \checkmark & & & P(X)=\square \\ \hline not X & & \checkmark & & \checkmark & \checkmark & P(\operatorname{not} X)=\square \\ \hline \end{tabular}

STEP 8

Calculate the probability of event $X$ , $P(X)$ , by dividing the number of white slices by the total number of slices.

STEP 9

Since there are 2 white slices and 5 total slices, the probability of event $X$ is:
$P(X) = \frac{\text{Number of white slices}}{\text{Total number of slices}} = \frac{2}{5}$

STEP 10

Fill in the probability of event $X$ in the table.
\begin{tabular}{|c|c|c|c|c|c|c|} \hline \multirow{2}{}{ Event } & \multicolumn{5}{|c|}{ Outcomes } & \multirow{2}{}{ Probability } \\ \cline { 2 - 6 } & 1 & 2 & 3 & 4 & 5 & \\ \hlineX & \checkmark & & \checkmark & & & P(X)=\frac{2}{5} \\ \hline not X & & \checkmark & & \checkmark & \checkmark & P(\operatorname{not} X)=\square \\ \hline \end{tabular}

STEP 11

Calculate the probability of event not $X$ , $P($ not $X)$ , by dividing the number of grey slices by the total number of slices.

STEP 12

Since there are 3 grey slices and 5 total slices, the probability of event not $X$ is:
$P(\text{not } X) = \frac{\text{Number of grey slices}}{\text{Total number of slices}} = \frac{3}{5}$

STEP 13

Fill in the probability of event not $X$ in the table.
\begin{tabular}{|c|c|c|c|c|c|c|} \hline \multirow{2}{}{ Event } & \multicolumn{5}{|c|}{ Outcomes } & \multirow{2}{}{ Probability } \\ \cline { 2 - 6 } & 1 & 2 & 3 & 4 & 5 & \\ \hlineX & \checkmark & & \checkmark & & & P(X)=\frac{2}{5} \\ \hline not X & & \checkmark & & \checkmark & \checkmark & P(\operatorname{not} X)=\frac{3}{5} \\ \hline \end{tabular}
The probabilities for the events are $P(X) = \frac{2}{5}$ and $P(\text{not } X) = \frac{3}{5}$ .

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Question A wheel has 5 slices numbered 1 to 5, some grey and some white. Find $P(X)$ , the probability the wheel stops on a white slice, and $P(\text{not }X)$ , the probability it stops on a non-white slice.

Studdy Solution

STEP 1

STEP 2

To determine the outcomes contained in event $X$ , identify the slices that are white.

STEP 3

Since slices 1 and 3 are white, these are the outcomes for event $X$ .

STEP 4

STEP 5

To determine the outcomes contained in event not $X$ , identify the slices that are not white, which are grey.

STEP 6

Since slices 2, 4, and 5 are grey, these are the outcomes for event not $X$ .

STEP 7

STEP 8

Calculate the probability of event $X$ , $P(X)$ , by dividing the number of white slices by the total number of slices.

STEP 9

Since there are 2 white slices and 5 total slices, the probability of event $X$ is:
$P(X) = \frac{\text{Number of white slices}}{\text{Total number of slices}} = \frac{2}{5}$

STEP 10

STEP 11

Calculate the probability of event not $X$ , $P($ not $X)$ , by dividing the number of grey slices by the total number of slices.

STEP 12

Since there are 3 grey slices and 5 total slices, the probability of event not $X$ is:
$P(\text{not } X) = \frac{\text{Number of grey slices}}{\text{Total number of slices}} = \frac{3}{5}$

STEP 13

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Question A wheel has 5 slices numbered 1 to 5, some grey and some white. Find P(X)P(X)P(X), the probability the wheel stops on a white slice, and P(not X)P(\text{not }X)P(not X), the probability it stops on a non-white slice.

Studdy Solution

STEP 1

STEP 2

To determine the outcomes contained in event XXX, identify the slices that are white.

STEP 3

Since slices 1 and 3 are white, these are the outcomes for event XXX.

STEP 4

STEP 5

To determine the outcomes contained in event not XXX, identify the slices that are not white, which are grey.

STEP 6

Since slices 2, 4, and 5 are grey, these are the outcomes for event not XXX.

STEP 7

STEP 8

Calculate the probability of event XXX, P(X)P(X)P(X), by dividing the number of white slices by the total number of slices.

STEP 9

Since there are 2 white slices and 5 total slices, the probability of event XXX is:P(X)=Number of white slicesTotal number of slices=25P(X) = \frac{\text{Number of white slices}}{\text{Total number of slices}} = \frac{2}{5}P(X)=Total number of slicesNumber of white slices​=52​

STEP 10

STEP 11

Calculate the probability of event not XXX, P(P(P( not X)X)X), by dividing the number of grey slices by the total number of slices.

STEP 12

STEP 13

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Question A wheel has 5 slices numbered 1 to 5, some grey and some white. Find $P(X)$ , the probability the wheel stops on a white slice, and $P(\text{not }X)$ , the probability it stops on a non-white slice.

To determine the outcomes contained in event $X$ , identify the slices that are white.

Since slices 1 and 3 are white, these are the outcomes for event $X$ .

To determine the outcomes contained in event not $X$ , identify the slices that are not white, which are grey.

Since slices 2, 4, and 5 are grey, these are the outcomes for event not $X$ .

Calculate the probability of event $X$ , $P(X)$ , by dividing the number of white slices by the total number of slices.

Since there are 2 white slices and 5 total slices, the probability of event $X$ is:
$P(X) = \frac{\text{Number of white slices}}{\text{Total number of slices}} = \frac{2}{5}$

Calculate the probability of event not $X$ , $P($ not $X)$ , by dividing the number of grey slices by the total number of slices.