Questions: A dartboard has 5 equally sized slices numbered from 1 to 5. Some are grey and some are white. The slices numbered 1,2,4, and 5 are grey. The slice numbered 3 is white. A dart is tossed and lands on a slice at random. Let X be the event that the dart lands on a grey slice, and let P(X) be the probability of X. Let not X be the event that the dart lands on a slice that is not grey, and let P( not X) be the probability of not X. (a) For each event in the table, check the outcome(s) that are contained in the event. Then, in the last column, enter the probability of the event. Event Outcomes Probability ----- 1 2 3 4 5 X P(x)= not X P(not X)= (b) Subtract. 1-P(x)=

A dartboard has 5 equally sized slices numbered from 1 to 5. Some are grey and some are white. The slices numbered 1,2,4, and 5 are grey. The slice numbered 3 is white. A dart is tossed and lands on a slice at random. Let X be the event that the dart lands on a grey slice, and let P(X) be the probability of X. Let not X be the event that the dart lands on a slice that is not grey, and let P( not X) be the probability of not X. (a) For each event in the table, check the outcome(s) that are contained in the event. Then, in the last column, enter the probability of the event. Event Outcomes Probability ----- 1 2 3 4 5 X P(x)= not X P(not X)= (b) Subtract. 1-P(x)=
Transcript text: A dartboard has 5 equally sized slices numbered from 1 to 5. Some are grey and some are white. The slices numbered 1,2,4, and 5 are grey. The slice numbered 3 is white. A dart is tossed and lands on a slice at random. Let $X$ be the event that the dart lands on a grey slice, and let $P(X)$ be the probability of $X$. Let not $X$ be the event that the dart lands on a slice that is not grey, and let $P($ not $X)$ be the probability of not $X$. (a) For each event in the table, check the outcome(s) that are contained in the event. Then, in the last column, enter the probability of the event. \begin{tabular}{|c|c|c|c|c|c|c|c|} \hline \multirow{2}{*}{ Event } & \multicolumn{5}{|c|}{ Outcomes } & \multirow{2}{*}{ Probability } \\ \cline { 2 - 5 } & 1 & 2 & 3 & 4 & 5 & \\ \hline$X$ & $\square$ & $\square$ & $\square$ & $\square$ & $\square$ & & $P(x)=\square$ \\ \hline not $X$ & $\square$ & $\square$ & $\square$ & $\square$ & $\square$ & $P(\operatorname{not} X)=\square$ \\ \hline \end{tabular} (b) Subtract. \[ 1-P(x)= \] $\square$
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Solution

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Solution Steps

Step 1: Identify the Events and Outcomes

The dartboard has 5 slices numbered from 1 to 5. The slices numbered 1, 2, 4, and 5 are grey, and the slice numbered 3 is white. We define the events as follows:

  • Event \(X\): The dart lands on a grey slice. The outcomes for this event are slices 1, 2, 4, and 5.
  • Event not \(X\): The dart lands on a slice that is not grey. The outcome for this event is slice 3.
Step 2: Calculate the Probability of Event \(X\)

The probability of event \(X\), denoted as \(P(X)\), is the number of grey slices divided by the total number of slices. There are 4 grey slices out of 5 total slices.

\[ P(X) = \frac{4}{5} = 0.8 \]

Step 3: Calculate the Probability of Event not \(X\)

The probability of event not \(X\), denoted as \(P(\text{not } X)\), is the number of non-grey slices divided by the total number of slices. There is 1 non-grey slice out of 5 total slices.

\[ P(\text{not } X) = \frac{1}{5} = 0.2 \]

Step 4: Subtract \(1 - P(X)\)

To find \(1 - P(X)\), we subtract the probability of event \(X\) from 1.

\[ 1 - P(X) = 1 - 0.8 = 0.2 \]

Final Answer

  • For event \(X\), the outcomes are slices 1, 2, 4, and 5, and the probability is \(\boxed{0.8}\).
  • For event not \(X\), the outcome is slice 3, and the probability is \(\boxed{0.2}\).
  • The result of \(1 - P(X)\) is \(\boxed{0.2}\).
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