9. frozen yogurt kali has a choice of 20 flavors for her triple scoop cone. if she chooses the flavors at…

Explanation:

Step1: Determine the total number of ways to choose 3 flavors from 20 (permutations, since order might matter? Wait, actually, for a triple scoop, does the order of scoops matter? Wait, the problem says "chooses the flavors at random" – if we consider that the three flavors are distinct and we are choosing 3 distinct flavors, the total number of ordered triples (permutations) would be ( P(20, 3)=\frac{20!}{(20 - 3)!}=20\times19\times18). Alternatively, if order doesn't matter, it's combinations ( C(20, 3)=\frac{20!}{3!(20 - 3)!}). But let's check the favorable outcome: the number of ways to have vanilla, chocolate, strawberry. If order matters, the number of favorable permutations is ( 3! = 6) (since the three flavors can be arranged in 3! ways). If order doesn't matter, the number of favorable combinations is 1 (since we want exactly those three flavors). Wait, the problem says "the 3 flavors she chooses will be vanilla, chocolate, and strawberry" – so maybe order doesn't matter, so we use combinations.

Wait, let's re-examine. The problem is about a triple scoop cone. So each scoop is a flavor. So if we consider the three scoops as positions (first, second, third), then the total number of possible ordered triples (since the order of scoops might matter, e.g., first scoop vanilla, second chocolate, third strawberry is different from first chocolate, second vanilla, third strawberry) – so permutations.

Total number of ordered triples (permutations) of 20 flavors taken 3 at a time: ( P(20, 3)=20\times19\times18 = 6840).

Favorable outcomes: the number of ordered triples that are vanilla, chocolate, strawberry. The three flavors can be arranged in ( 3! = 6) ways (since each of the three positions can have any of the three flavors, as long as all three are present). Wait, no: the favorable outcome is that the three flavors are exactly vanilla, chocolate, strawberry, in any order. So the number of favorable permutations is the number of permutations of 3 distinct items, which is ( 3! = 6).

Wait, but maybe the problem is considering that the three flavors are chosen without regard to order (i.e., combinations), because a "choice of flavors" might be a set. Let's check both approaches.

First, using permutations (order matters):

Total permutations: ( 20\times19\times18 = 6840).

Favorable permutations: number of ways to arrange vanilla, chocolate, strawberry in the three scoops: ( 3! = 6).

Then probability would be ( \frac{6}{6840}=\frac{1}{1140}\approx0.000877) (which is about 0.0877%). But that doesn't match the handwritten note. Wait, the handwritten note has ( 20\times19\times18 = 6840)? Wait, no, 20×19×18 is 6840, but the handwritten note has 20×19×18? Wait, the handwritten note says "20·19·18 = 6840" (wait, 20×19×18 is 6840, not 7280 – maybe a typo in the handwritten note). Then the favorable outcome: if we consider that the three flavors must be vanilla, chocolate, strawberry in any order, but the problem says "the 3 flavors she chooses will be vanilla, chocolate, and strawberry" – maybe the order doesn't matter, so the number of favorable combinations is 1, and total combinations is ( C(20, 3)=\frac{20!}{3!17!}=\frac{20\times19\times18}{6}=1140). Then the probability would be ( \frac{1}{1140}\approx0.000877), but that's the same as ( \frac{6}{6840}) (since ( 6840 = 6\times1140), so ( \frac{6}{6840}=\frac{1}{1140})).

Wait, the handwritten note has "20·19·18 = 7280" – that's a mistake, 20×19×18 is 6840. Then "1/7280" – no, that's wrong. Wait, maybe the problem is that the three flavors are chosen with replacement? No, because it's a triple scoop with three flavors (vanilla, chocolate, strawberry – three distinct flavors), so replacement is not allowed.

Wait, let's start over.

Step 1: Determine the total number of possible outcomes.

If we consider that the three scoops are distinct (i.e., the order of flavors matters, like first scoop, second scoop, third scoop), then the total number of possible flavor combinations (with distinct flavors) is the number of permutations of 20 flavors taken 3 at a time:

( P(20, 3) = 20 \times 19 \times 18 = 6840 )

Step 2: Determine the number of favorable outcomes.

The favorable outcome is that the three scoops are vanilla, chocolate, and strawberry, in any order. The number of ways to arrange three distinct flavors in three positions is ( 3! = 6 ) (since for the first scoop, we can choose any of the 3 flavors, the second scoop any of the remaining 2, and the third scoop the last 1: ( 3 \times 2 \times 1 = 6 )).

Step 3: Calculate the probability.

Probability ( P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3!}{P(20, 3)} = \frac{6}{6840} = \frac{1}{1140} \approx 0.000877 ) (or approximately 0.0877%).

Alternatively, if we consider that the order of flavors does not matter (i.e., combinations), then:

Total number of combinations: ( C(20, 3) = \frac{20!}{3!(20 - 3)!} = \frac{20 \times 19 \times 18}{6} = 1140 )

Number of favorable combinations: 1 (since we want exactly the three flavors vanilla, chocolate, strawberry, regardless of order)

Then probability ( P = \frac{1}{1140} \approx 0.000877 ), which is the same as before.

Wait, the handwritten note has "20·19·18 = 7280" – that's an error, 20×19=380, 380×18=6840. Then "1/7280" – no, that's wrong. But let's check the correct calculation.

So the correct probability is ( \frac{1}{C(20, 3)} ) if order doesn't matter, or ( \frac{3!}{P(20, 3)} ) if order matters. Both give ( \frac{1}{1140} \approx 0.000877 ), or approximately 0.0877%.

Wait, but maybe the problem is considering that the three flavors are chosen without replacement and order doesn't matter, so:

Total number of ways to choose 3 flavors from 20: ( C(20, 3) = \frac{20!}{3!17!} = \frac{20 \times 19 \times 18}{6} = 1140 )

Number of favorable ways: 1 (since we want exactly those three flavors)

Thus, probability ( P = \frac{1}{1140} \approx 0.000877 ), or ( \frac{1}{1140} \approx 0.0877% ).

Alternatively, if order matters, total ways: ( 20 \times 19 \times 18 = 6840 ), favorable ways: ( 3! = 6 ), so ( P = \frac{6}{6840} = \frac{1}{1140} ), same result.

So the correct probability is ( \frac{1}{1140} ) or approximately 0.000877 (or 0.0877%).

Answer:

The probability is (\boxed{\dfrac{1}{1140}}) (or approximately (0.000877) or (0.0877%)).