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tester [92]
2 years ago
10

Two married couples have purchased theater tickets and are seated in a row consisting of just 4 seats. If they take their seats

in a completely random order, what is the probability that at least one of the wives ends up sitting next to her husband? Hint: Use probability of a complement event. That is, P(A) = 1 − P(A′), where A′ is the complement of an event A.
Mathematics
1 answer:
Licemer1 [7]2 years ago
7 0

Answer:

The answer is 0.75.

Step-by-step explanation:

To use probability of a complement event, we need to find the situation where none of the couples are sitting next to each other which can happen either two husbands are sitting in the seats 2 and 3 or two wives are sitting in the seats 2 and 3. Also in the first scenario, the wives would need to be sitting at the opposite ends to their husbands and vice versa for the second scenario.

So there are in total 4 scenarios where no couple is sitting next to each other.

There are a total of 16 different ways that they can be seated so the probability of compelement is 16 - 4 = 12, which is 3/4 of the total, meaning that the probability of at least one of the wives sittting next to her husband is 0.75 or 75%.

I hope this answer helps.

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irinina [24]

Answer: 3/4 as a decimal is 0.75


Step-by-step explanation: 0.75 is the percent. so it is 75%

Hope this helps :)

4 0
3 years ago
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One of the assumptions underlying the theory of control charting (see Chapter 16) is that successive plotted points are independ
Illusion [34]

Question has missing details

One of the assumptions underlying the theory of control charting (see Chapter 16) is that successive plotted points are independent of one another. Each plotted point can signal either that a manufacturing process is operating correctly or that there is some sort of malfunction. Even when a process is running correctly, there is a small probability that a particular point will signal a problem with the process. Suppose that this probability is 0.05. What is the probability that at least one of 10 successive points indicates a problem when in fact the process is operating correctly?

Answer:

The probability that at least one of 10 successive points indicates a problem when in fact the process is operating is 0.4013

Step-by-step explanation:

Given

Let P = Probability that a point signals an error incorrectly = 0.05

Let Q = Probability that a point signals an error correctly

P + Q = 1 ---- Make Q the subject of formula

Q = 1 - P where P = 0.05

So, Q = 1 - P becomes

Q = 1 - 0.05

Q= 0.95

Solving for the probability that at least one of 10 successive points indicates a problem when in fact the process is operating.

If two events (P and Q) are independent

Then

P(P n Q) = P(P) * P(Q)

From De Morgan law;

P(P u Q) = 1 - P(P' n Q')

Where P(P u Q) represent the probability that at least one of 10 successive points

P(P' n Q') is calculated as follows;

P(P' n Q') = 0.95^10

P(P' n Q') = 0.59873693923837890625

So,

P(P u Q) = 1 - P(P' n Q') becomes

P(P u Q) = 1 - 0.59873693923837890625

P(P u Q) = 0.40126306076162109375

P(P u Q) = 0.4013 ----- Approximated

The probability that at least one of 10 successive points indicates a problem when in fact the process is operating is 0.4013

7 0
3 years ago
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Vesnalui [34]

Answer:

About 8.6

Step-by-step explanation:

d=(x2-x1)^2+(y2-y1)^2

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(3-8)^2+(1-8)^2

(-5)^2+(-7)^2

25+49

74

take the square root and you get about 8.6

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Answer:

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Step-by-step explanation:

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Answer:

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