Question 206164This question is from textbook Essential Statistics in Business and Economics
: Bob, Mary, and Jen go to dinner. Each orders a different meal. The waiter forgets who ordered which meal, so he randomly places the meals before the three diners. Let C be the event that a diner gets the correct meal and let N be the event that a diner gets an incorrect meal. Enumerate the sample space and then find the probability that:
a. No diner gets the correct meal
b. Exactly one diner gets the correct meal
c. Exactly two diners get the correct meal
d. All three diners get the correct meal
This question is from textbook Essential Statistics in Business and Economics
Found 2 solutions by stanbon, waddylemons:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Bob, Mary, and Jen go to dinner. Each orders a different meal. The waiter forgets who ordered which meal, so he randomly places the meals before the three diners. Let C be the event that a diner gets the correct meal and let N be the event that a diner gets an incorrect meal. Enumerate the sample space and then find the probability that:
Sample space: 1 means got their order ; o means did not get their order
000,001,010,011,100,101,110,111
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a. No diner gets the correct meal-------------------1/8
b. Exactly one diner gets the correct meal----------3/8
c. Exactly two diners get the correct meal----------3/8
d. All three diners get the correct meal------------1/8
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Cheers,
Stan H.
Answer by waddylemons(1) (Show Source):
You can put this solution on YOUR website! To find the answer to this problem, you first have to make a list of all the possible outcomes. Let's say Bob's meal is A, Mary's meal is B, and Jen's meal is C. The 6 possible outcomes for the order in which the waiter hands out their meals are as follows:
ABC
ACB
BAC
BCA
CAB
CBA
If we let "ABC" represent the outcome where all 3 diners receive the correct meal, we can calculate the probability for the other outcomes (nobody gets the right meal, 1/3 get the correct meal, 2/3 get the correct meal).
ABC ... (everyone gets the correct meal)
ACB ... (only Bob gets the correct meal)
BAC ... (only Jen gets the correct meal)
BCA ... (nobody gets the correct meal)
CAB ... (nobody gets the correct meal)
CBA ... (only Mary gets the correct meal)
As you can see, it's not possible for only 2 people to receive the correct meal (if someone has the wrong meal then that means they received someone else's meal...making at least 2/3 meals wrong).
We were asked to calculate the probability that:
A. No diner gets the correct meal (2/6 = 33.3%)
B. Exactly one diner gets the correct meal (3/6 = 50%)
C. Exactly two diners get the correct meal (0/6 = 0.00%)
D. All diners get the correct meal (1/6 = 16.7%)
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