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28 Nov 2023, 08:05
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Number of lots: 20
Desirable outcomes in B: 16
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
82% (02:13) correct
18%
(02:19)
wrong
based on 629
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Data Insights (DI) Butler 2023-24 [Question #132, Date: Nov-28-2023] [Click here for Details]
During a lottery, several lots are to be selected. A is the event that a certain subset of these lots is selected, and B is the event that another subset of these lots is selected, such that:
• the probability of event A occurring is 2/5.
• the probability of event B occurring is 4/5.
• the probability of the union of events A and B occurring is 1.
• the intersection of events A and B is an event with four desirable outcomes.
In the following table, identify the total number of lots and the number of desirable outcomes in event B. Make only one selection in each column.
| Number of lots | Desirable outcomes in B | |
| 15 | ||
| 16 | ||
| 17 | ||
| 18 | ||
| 19 | ||
| 20 |
ShowHide Answer
Official Answer
Number of lots: 20
Desirable outcomes in B: 16
01 Dec 2023, 08:12
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Official Explanation
The probability of the union of events A and B occurring equals the probability of event A occurring, plus the probability of event B occurring, minus the probability of the intersection of events A and B occurring. You know the value of all of these probabilities except the last one. The probability of the intersection of events A and B equals the number of desirable outcomes in the intersection of A and B, which is 4, over the number of total outcomes in the lottery—let’s call that t. Thus, you have:
P (AUB) = P (A) + P (B) - P (A∩B)
\(1=\frac{2}{5}+\frac{4}{5}-\frac{4}{t}\)
\(1=\frac{2t-4t-20}{5t}\)
\(5t=6t-20\)
\(t=20\)
So the correct answer in the first column = the total number of lots is 20.
Next, let the number of desirable outcomes in event B be b. In that case, you have:
P(B)\(=\frac{b}{20}\)
\(\frac{4}{5}=\frac{b}{20}\)
\(4=\frac{b}{4}\)
\(b=16\)
So the correct answer in the second column is: the number of desirable outcomes in event B is 16.
The probability of the union of events A and B occurring equals the probability of event A occurring, plus the probability of event B occurring, minus the probability of the intersection of events A and B occurring. You know the value of all of these probabilities except the last one. The probability of the intersection of events A and B equals the number of desirable outcomes in the intersection of A and B, which is 4, over the number of total outcomes in the lottery—let’s call that t. Thus, you have:
P (AUB) = P (A) + P (B) - P (A∩B)
\(1=\frac{2}{5}+\frac{4}{5}-\frac{4}{t}\)
\(1=\frac{2t-4t-20}{5t}\)
\(5t=6t-20\)
\(t=20\)
So the correct answer in the first column = the total number of lots is 20.
Next, let the number of desirable outcomes in event B be b. In that case, you have:
P(B)\(=\frac{b}{20}\)
\(\frac{4}{5}=\frac{b}{20}\)
\(4=\frac{b}{4}\)
\(b=16\)
So the correct answer in the second column is: the number of desirable outcomes in event B is 16.
General Discussion
29 Nov 2023, 08:41
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hi Sajjad1994, could you help to explain this question?
