Question 1142628
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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.<br>
So, assuming all three daily doubles are found....<br>
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<br>
P(all three by one contestant): (1)(1/3)(1/3) = 1/9<br>
b. P(all three by the returning champion)...<br>
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<br>
P(all three by the returning champion) = (1/3)(1/3)(1/3) = 1/27<br>
c. P(one by each of the three contestants)...<br>
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<br>
P(one by each of the three contestants) = (1)(2/3)(1/3) = 2/9