A travel writer is on a cruise. After the cruise, he is going to write
a magazine article evaluating the quality of the cruise, including
the food. (For instance, "On the first night, I had the shark steak
Florentine ... which was superb.") The dinner menu, which
is the same every evening, has 8 fish entrees and 9 non-fish
entrees (e.g., beef, pork, etc.). Each night the writer
selects a different entree and records the selection in his
notebook. The cruise will last 6 nights, so there will be 6 dinnner
entrees recorded.
How many overall outcomes are possible?
In how many possible overall outcomes would fish entrees appear
exactly 4 times?
In how many possible overall outcomes would fish entrees appear at most
4 times?
In how many possible overall outcomes would fish entrees appear at least
1 time?
How many overall outcomes are possible? -- C(17,6) = 12376
In how many possible overall outcomes would fish entrees appear exactly 4 times?
C(8,4)xC(9,2) = 2520
In how many possible overall outcomes would fish entrees appear at least 1 time?
we must exclude the no of cases of no-fish, which would be C(9,6) or 84
so no of cases with at least 1 fish dinner is 12376-84 = 12292
To answer these questions, we can use the concept of combinations and the principle of multiplication.
1. How many overall outcomes are possible?
For each of the 6 nights, the travel writer can choose from 8 fish entrees and 9 non-fish entrees. Since the choices are independent, we multiply the number of choices for each night. Hence, the total number of overall outcomes is 8 * 9 * 8 * 9 * 8 * 9 = 48,384.
2. In how many possible overall outcomes would fish entrees appear exactly 4 times?
To calculate this, we need to consider the number of ways to choose 4 nights for fish entrees out of the 6 nights and multiply it by the number of ways to choose non-fish entrees for the remaining 2 nights. This can be calculated using combinations. The number of ways to choose 4 nights out of 6 for fish entrees is C(6, 4) = 15. For the remaining 2 nights, the travel writer can choose from non-fish entrees, giving us 9 * 9 = 81 choices. Thus, the total number of outcomes with fish entrees appearing exactly 4 times is 15 * 81 = 1,215.
3. In how many possible overall outcomes would fish entrees appear at most 4 times?
We need to calculate the sum of outcomes where fish entrees appear exactly 0, 1, 2, 3, or 4 times. We already know the number of outcomes where fish entrees appear exactly 4 times (1,215) from the previous question. For 0, 1, 2, and 3 times, we can use similar combinations.
For 0 times: C(6, 0) * 9^6 = 9^6 = 531,441 (no fish entrees on any night).
For 1 time: C(6, 1) * (8 * 9^5) = 6 * 8 * 9^5 = 2,466,624.
For 2 times: C(6, 2) * (8^2 * 9^4) = 15 * 8^2 * 9^4 = 3,724,416.
For 3 times: C(6, 3) * (8^3 * 9^3) = 20 * 8^3 * 9^3 = 1,342,208.
Summing up these results, we have 531,441 + 2,466,624 + 3,724,416 + 1,342,208 + 1,215 = 9,066,904 possible outcomes with fish entrees appearing at most 4 times.
4. In how many possible overall outcomes would fish entrees appear at least 1 time?
To calculate this, we subtract the outcome where fish entrees don't appear at all (531,441) from the total number of possible outcomes (48,384). Therefore, there are 48,384 - 531,441 = 47,853 possible outcomes with fish entrees appearing at least 1 time.