An ice chest contains 5 cans of apple juice, 6 cans of grape juice, 6 cans of orange juice,
and 4 cans of mango juice. Suppose that you reach into the container and randomly select
three cans in succession. Find the probability of selecting three cans of grape juice.
The probability of selecting from the ice chest three cans of grape juice is
(Type an integer or a simplified fraction.)

Question

UpStudy AI · Accepted Answer

We start by finding the total number of cans in the ice chest:

$$
\text{Total cans} = 5 + 6 + 6 + 4 = 21.
$$

Since there are 6 cans of grape juice, the probability that the first can selected is grape juice is:

$$
\frac{6}{21}.
$$

After selecting one grape juice can, there are 5 grape juice cans left and a total of 20 cans remaining. Thus, the probability that the second can is grape juice is:

$$
\frac{5}{20}.
$$

After selecting a second grape juice can, 4 grape juice cans remain and the total number of cans is now 19. Therefore, the probability that the third can is grape juice is:

$$
\frac{4}{19}.
$$

The overall probability of selecting three cans of grape juice in succession is the product of the individual probabilities:

$$
\frac{6}{21} \times \frac{5}{20} \times \frac{4}{19} = \frac{6 \times 5 \times 4}{21 \times 20 \times 19} = \frac{120}{7980}.
$$

To simplify this fraction, we factor both the numerator and the denominator:

- Numerator: $$120 = 2^3 \times 3 \times 5$$.
- Denominator: $$21 \times 20 \times 19 = (3 \times 7) \times (2^2 \times 5) \times 19 = 2^2 \times 3 \times 5 \times 7 \times 19$$.

Cancel the common factors $$2^2$$, $$3$$, and $$5$$:

$$
\frac{2^3 \times 3 \times 5}{2^2 \times 3 \times 5 \times 7 \times 19} = \frac{2}{7 \times 19} = \frac{2}{133}.
$$

Thus, the probability of selecting three cans of grape juice is:

$$
\frac{2}{133}.
$$
The probability of selecting three cans of grape juice is $$\frac{2}{133}$$.

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