Three married couples have purchased theater tickets and are seated in a row con
ID: 3205933 • Letter: T
Question
Three married couples have purchased theater tickets and are seated in a row consisting of just six seats. If they take their seats in a completely random order…
What is the probability that Bob and Jackie (who are husband and wife) sit in the two seats on the far left?
What is the probability that Bob and Jackie end up sitting next to one another?
What is the probability that at least one of the wives ends up sitting next to her husband?
Now use the multiplication rule to compute the probability that Bob and Jackie sit together on the far left (event A) and that Mike and Eve (husband and wife) sit together in the middle (event B).
Given that Mike and Eve sit together in the middle, what is the probability that the two other husbands sit next to their wives?
Given that Mike and Eve sit together, what is the probability that all husbands sit next to their wives?
Explanation / Answer
probability that Bob and Jackie (who are husband and wife) sit in the two seats on the far left = 2*4!/6! = 1/15
probability that at least one of the wives ends up sitting next to her husband = 3C1 *5!/6! = 1/2
probability that Bob and Jackie sit together on the far left (event A) and that Mike and Eve (husband and wife) sit together in the middle (event B) =P(A)*P(B|A) = 4/15
Given that Mike and Eve sit together in the middle, what is the probability that the two other husbands sit next to their wives = 8/2*4! = 1/3! =1/6
Given that Mike and Eve sit together, what is the probability that all husbands sit next to their wives = 8*3!/5!
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