What's the probability of a household subscribing to at least one newspaper?

The probability that a random household subscribes to at least one paper is 90/100= 0.9. The probability that a random household subscribes to exactly one paper is 40/100= 0.4.In summary, the probability that a random household subscribes to at least one newspaper is 0.9 and the probability that a random household subscribes to exactly one newspaper is 0.4.
  • #1
caljuice
70
0

Homework Statement



If 60% of households subscribe to Metro(M) newspaper, 80% subscribe to local (L) newspaper, and 50% subscribe to both,

1)what's the probability that a random household subscribes to at least one paper?
2) what's the probability that a random household subscribes to exactly one paper?

The Attempt at a Solution



1) The probability of at least one paper subscribed is P(M U L) = P(M) + P(L) - P(M AND L) which gives the answer of 0.9. But I'm wondering why this works? The phrasing "at least" means the possibilities can be just P(M), just P(L), or P(M AND L) but from the equation we are substracting out P(M AND L), which means P(M) and P(L) are the only possible outcomes. What's wrong with my thinking here?

2) Probability = P(Mc and L) U P(Mc and L) and they are mutually exclusive so you just sum the two probabilities.

Pc = 1- P
Not sure how to find P(Mc and L) since they aren't independent.

thanks.
 
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  • #2
caljuice said:

Homework Statement



If 60% of households subscribe to Metro(M) newspaper, 80% subscribe to local (L) newspaper, and 50% subscribe to both,

1)what's the probability that a random household subscribes to at least one paper?
2) what's the probability that a random household subscribes to exactly one paper?


The Attempt at a Solution



1) The probability of at least one paper subscribed is P(M U L) = P(M) + P(L) - P(M AND L) which gives the answer of 0.9. But I'm wondering why this works? The phrasing "at least" means the possibilities can be just P(M), just P(L), or P(M AND L) but from the equation we are substracting out P(M AND L), which means P(M) and P(L) are the only possible outcomes. What's wrong with my thinking here?

2) Probability = P(Mc and L) U P(Mc and L) and they are mutually exclusive so you just sum the two probabilities.

Pc = 1- P
Not sure how to find P(Mc and L) since they aren't independent.

thanks.

P(L & M) is part of P(M) and is also part of P(L), so the summation P(L) + P(M) counts P(L&M) twice. That is why you need to subtract it---so you only count it once.

RGV
 
  • #3
Imagine that there are 100 households. Then 60 subscribe to M and 50 subscribe to both so 10 subscribe to M only. 80 subscribe to L and 50 subscribe to both so 30 subscribe to L only. That makes a total of 10+ 50+ 30= 90 households out of 100 that subscribe to at least one newspaper and 10+ 30= 40 that subscribe to exactly one newspaper.
 

Related to What's the probability of a household subscribing to at least one newspaper?

1. What is probability union and how is it used?

Probability union is a mathematical concept that refers to the likelihood of two or more events occurring together. It is used to calculate the probability of a union of events, which is the probability that at least one of the events will occur.

2. How is probability union different from probability intersection?

Probability union and probability intersection are two different concepts in probability. Probability intersection refers to the likelihood of two or more events occurring together, while probability union refers to the likelihood of at least one of the events occurring. In other words, probability intersection looks at the overlap between events, while probability union looks at the combined probability of events.

3. What is the formula for calculating probability union?

The formula for calculating probability union is P(A ∪ B) = P(A) + P(B) - P(A ∩ B), where P(A) and P(B) are the individual probabilities of events A and B, and P(A ∩ B) is the probability of the intersection of events A and B.

4. How can probability union be applied in real life?

Probability union can be applied in various real-life scenarios, such as in risk assessment, insurance, and decision-making. For example, in risk assessment, probability union can be used to calculate the likelihood of multiple hazards occurring simultaneously, while in insurance, it can be used to determine the probability of multiple events that may result in a claim. In decision-making, probability union can help assess the likelihood of different outcomes and inform decision-making processes.

5. Are there any limitations to using probability union?

While probability union is a useful tool in probability theory, it does have some limitations. It assumes that events are independent of each other, which may not always be the case in real-life situations. Additionally, it can only be used for a finite number of events, and it may become more complex when dealing with a large number of events. It is important to carefully consider these limitations when using probability union in any application.

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