SOLUTION: There are three clues labeled “daily double” on the game show Jeopardy. If three
equally matched contenders play, what is the probability that:
a. A single contestant finds a
Algebra ->
Probability-and-statistics
-> SOLUTION: There are three clues labeled “daily double” on the game show Jeopardy. If three
equally matched contenders play, what is the probability that:
a. A single contestant finds a
Log On
Question 1142628: There are three clues labeled “daily double” on the game show Jeopardy. If three
equally matched contenders play, what is the probability that:
a. A single contestant finds all three “daily doubles”?
b. The returning champion gets all three of the “daily doubles”?
c. Each of the players selects precisely one of the “daily doubles”? Found 2 solutions by Alan3354, greenestamps:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! There are three clues labeled “daily double” on the game show Jeopardy. If three
equally matched contenders play, what is the probability that:
a. A single contestant finds all three “daily doubles”?
b. The returning champion gets all three of the “daily doubles”?
c. Each of the players selects precisely one of the “daily doubles”?
----------
You have to specify the total number of clues.
I think there are 30, but I haven't seen that show in decades.
I find it irritating.
----
Whenever you see "... we have to assume..." it's not longer a math problem.
----
assume = Ass + U + Me
The other tutor is technically right; the statement of the problem is not complete enough to answer the questions. To solve the problem, we have to assume that each of the three daily doubles is found by one of the contestants. On the actual game show, it is possible for one of the daily doubles not to be found before time expires.
So, assuming all three daily doubles are found....
a. P(all three by one contestant):
Any of the three can find the first one: P = 1
The same contestant has to find the second: P = 1/3
The same contestant also has to find the third: P = 1/3
P(all three by one contestant): (1)(1/3)(1/3) = 1/9
b. P(all three by the returning champion)...
The returning champion has to find the first one: P = 1/3
The returning champion also has to find the second: P = 1/3
The returning champion also has to find the third: P = 1/3
P(all three by the returning champion) = (1/3)(1/3)(1/3) = 1/27
c. P(one by each of the three contestants)...
Any of the three can find the first: P = 1
Either of the other two can find the second: P = 2/3
The third contestant has to find the third: P = 1/3
P(one by each of the three contestants) = (1)(2/3)(1/3) = 2/9
