The Social Security Administration documented the marriage year of various couples, along with the number of children they had. Use the table given below to answer the following. (... The Social Security Administration documented the marriage year of various couples, along with the number of children they had. Use the table given below to answer the following. (a) If a couple is selected at random, what is the probability the couple was married in 2000 and had two children? (b) If a couple is selected at random, what is the probability the couple was married before 1998 or had three children? (c) Compute P((B∪C)∩2). (d) Compute P((A∩E)∪3). (e) Compute P(E^C∩4). (f) Compute P((B∪D)^C∩1).
Understand the Problem
The image shows a table documenting the marriage year of various couples and the number of children they have. Several probability questions need to be answered using the provided table. The questions involve calculating probabilities based on specific events related to marriage year and number of children, as well as set operations on these events.
Answer
(b) 135/251 (f) 27/251
Answer for screen readers
(b) 135/251 (f) 27/251
Steps to Solve
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(b) Calculate the probability of marriage before 1998
Marriage before 1998 means the couples were married in 1996 or 1997. The total number of such couples is $38 + 56 = 94$.
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(b) Find the number of couples with three children
The total number of couples with three children is 70 from the table.
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(b) Identify the overlap
We need to subtract the couples that satisfy both conditions (married before 1998 AND have three children) to avoid double-counting. Couples married in 1996 and had three children: 9. Couples married in 1997 and had three children: 20. The total number of such couples: $9+20=29$.
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(b) Apply the inclusion-exclusion principle
The number of couples married before 1998 OR had three children is: (couples married before 1998) + (couples with three children) - (couples married before 1998 AND had three children) $= 94 + 70 - 29 = 135$.
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(b) Calculate the probability
The probability is the number of favorable outcomes divided by the total number of outcomes: $135/251$. 6. (f) Compute the couples in $B \cup D$
$B \cup D$ means couples married in 1997 or 1999. The number of such couples is $56 + 57 = 113$. 7. (f) Compute the couples in $(B \cup D)^C$
$(B \cup D)^C$ is the complement of $B \cup D$, so it contains all couples NOT in $B \cup D$. The total number of couples is 251, so the number of couples in $(B \cup D)^C$ is $251 - 113 = 138$. 8. (f) Compute the couples in $(B \cup D)^C \cap 1$
$(B \cup D)^C \cap 1$ means people not in $B \cup D$ AND have one child. The number of couples with one child is 50. $B$ has 17 couples with one child $D$ has 6 couples with one child Therefore, $B \cup D$ has $17 + 6 = 23$ couples with one child. So, $(B \cup D)^C \cap 1$ has $50 - 23 = 27$ couples. 9. (f) Calculate the probability
$P((B \cup D)^C \cap 1)$ is the number of couples in $(B \cup D)^C \cap 1$ divided by the total number of couples. The probability is therefore $27/251$.
(b) 135/251 (f) 27/251
More Information
Probabilities are always numbers between 0 and 1, inclusive, and can be expressed as fractions, decimals, or percentages. In this problem, we express the probabilities as fractions.
Tips
- For part (b), a common mistake is to add the number of couples married before 1998 and the number of couples with three children, but forget to subtract the intersection (couples that are both married before 1998 AND have three children). This is an application of the inclusion-exclusion principle.
- For part (f), a common mistake is misinterpreting the set operations and not correctly calculating the number of couples that satisfy the compound condition. Specifically, correctly finding the complement of the union and then finding the intersection with couples having one child.
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