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A restaurant has a total of 60 tables. Of those tables, 38 are round and 13 are located by the window. There are 6 round tables by the window.If tables are randomly assigned to customers, what is the probability that a customer will be seated at a round table or by the window? (29/60, 47/60, 41/60, or 45/60)

User Xiotee
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Given:


\begin{gathered} Total-table=60 \\ round-table=38 \\ window-side-table=13 \\ round-table(window-side)=6 \end{gathered}

To Determine: The probability that a customer will be seated at a round table or by the window

Solution

Probability is a ratio of the number of favorable outcomes to the number of possible outcomes of the experiment


Probability=(Number(outcome))/(Number(Possible-outcome))
\begin{gathered} P(round-table)=(38)/(60) \\ P(window)=(13)/(60) \\ P(round-window)=(6)/(60) \end{gathered}

The probability that a customer will be seated at a round table or by the window would be


\begin{gathered} P(round-table,or,window)=P(round)+P(window)-P(round-window) \\ =(38)/(60)+(13)/(60)-(6)/(60) \\ =(38+13-6)/(60) \\ =(51-6)/(60) \\ =(45)/(60) \end{gathered}

Hence, the probability that a customer will be seated at a round table or by the window is 45/60

User DerManu
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