Answer:
Question 1: The probability that a bus is owned by Jerry, given that it runs on time:
Question 2: The probability that a bus is owned by Jerry, given that it runs late
Explanation:
A two-way table is very good to deal with these kind of probability questions because helps you to determine the every subset.
I will guide you on how to build the two-way table.
Identify the possible sets. These are:
- Buses operated within a particular town (100)
- Buses that run on time
- Buses that run late
- Buses owned by Jerry
- Buses not owned by Jerry.
With that, you start to build your table:
<u>1. Start</u>
Run on time Run late Total
Owned by Jerry
Not owned by Jerry
Total 100
Now, lets fill using each statement:
<u>2. 48 usually run on time</u>
Run on time Run late Total
Owned by Jerry
Not owned by Jerry
Total 48 100
<u>3. 36 of these are owned by Jerry:</u>
Run on time Run late Total
Owned by Jerry 36
Not owned by Jerry
Total 48 100
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<u>4. Of the 52 buses that usually run behind, only 7 are owned by Jerry:</u>
Run on time Run late Total
Owned by Jerry 36 7 43
Not owned by Jerry
Total 48 52 100
You may even fill the second row, but you will not need it to answer the questions.
<u>5. Complete the table using differences:</u>
Run on time Run late Total
Owned by Jerry 36 7 43
Not owned by Jerry 12 45 57
Total 48 52 100
<u>6. Answer the questions:</u>
<u>Question 1</u>: The probability that a bus is owned by Jerry, given that it runs on time:
- Probability of an event = # favorable outcomes / # total outcomes
- Probability that a bus owned by Jerry, given that it runs late = # buses owned by Jerry that runs on time / # buses that run on time = 36 / 48 = 3/4
<u>Question 2</u>: The probability that a bus is owned by Jerry, given that it runs late
- Probability that a bus is owned by Jerry, given that it runs late = # buses owned by Jerry that run late / # buses that run late = 7 / 52
first. By doing this, it would then give us our answer as 16. By us understanding this point of view, we would then consider that this would then be your answer. That would then include that there would then be 16 pairs of the "enantiomeric pairs", and that would then be the possible estimate.