Question

A shop has 4 types of sweets (chocolate, taffy, gummies, and cookies), 2 types of snacks

Answer 1

Answer:

$\frac{1}{24}$

Step-by-step explanation:

Given: A shop has $4$ types of sweets (chocolate, taffy, gummies, and cookies), $2$ types of snacks  (chips and crackers), and $3$ types of drinks (sodas, juice, and sports drinks).  Mystery boxes are put together that randomly combine $1$ sweet, $1$ snack, and $1$ drink.

To Find: probability that a mystery box contains chocolate, chips, and juice.

Solution:

Let  A be the event that sweet in box is a chocolate.

      B be the event that snack in box is chips.

      C be the event that drink in box is juice.

shop has $4$ types of sweets chocolate, taffy, gummies, and cookies.

       $p(\text{A})=\frac{\text{desired outcomes}}{\text{possible outcome}}$

                       $=\frac{1}{4}$

shop has $2$ types of snacks  chips and crackers

        $p(\text{B})=\frac{\text{desired outcomes}}{\text{possible outcome}}$

                       $=\frac{1}{2}$

shop has $3$ types of drinks sodas, juice, and sports drinks.

        $p(\text{C})=\frac{\text{desired outcomes}}{\text{possible outcome}}$

                       $=\frac{1}{3}$

Now,

probability that box contains chocolate, chips and juice is

 $p(\text{A and B and C})$

using multiplication rule in probability

$p(\text{A and B and C})=p(\text{A}).p(\text{B}).p(\text{C})$

$p(\text{A and B and C})=\frac{1}{4}\times\frac{1}{2}\times\frac{1}{3}$

$p(\text{A and B and C})=\frac{1}{24}$              

Hence the probability that box contain chocolate, chis and juice is $\frac{1}{24}$