Question

A candy store grab bag contains 10 pieces of sour candy, 12 pieces of chocolate candy, and 6 pieces of sweet candy. If you select 1 piece of candy, replace it, and then select another piece, what is the probability that the first piece of candy taken out is a sour candy and the second piece of candy taken out is not a sour candy?
23%
92%
29%
100%

Answer 1

Answer:

A. 23%

Step-by-step explanation:

Let's figure out how many pieces of candy there is in total:

10 + 12 + 6 = 28

In total, there are 28 pieces of candy.

Let's find the probability that a piece is sour:

$\frac{10}{28} = \frac{5}{14}$

The probability of pulling a sour piece out of the bag is 5/14.

This means that the probability that the candy pulled is not sour is 9/15 (or simplified 3/5)

To find the probability that the first candy picked out is sour AND the second candy picked is not sour we have to multiply the probabilities of both:

$\frac{5}{14} *\frac{3}{5} = 0.229 = 23$

Therefore, A. 23% is the correct answer.

Hope this helps!!

- Kay :)

Answer 2

Step-by-step explanation:

so we're making two draws *with* replacement (this is important)

step 1: for the first draw, it wants the probability of getting a sour candy. to calculate this:

(# of sour candy) / (total # of candy)

step 2: for the second draw, it wants the probability of *not* getting a sour candy. to calculate this, you can calculate 1 - (the probability form part 1).

step 3: to find the probability of both events happening together, simply multiply the probabilities from part 1 and 2 together

side note: for step 2, you can only do this because the candy is being replaced. if there were no replacement, you'd have to re-calculate (# of non-sour candies) / (total after the first candy is drawn)