Question

A paper bag contains a mixture of 3 types of candy. there are 10 packs of fuzzy peaches, 7 packs of sour keys, and 3 packs of nerds. suppose a game is played in which a candy is randomly taken from the bag, replaced, and then a second candy is drawn from the bag. if you are allowed to keep the second candy only if it was the same type a the one that was drawn the first time, calculate the probability of each of the following: a) you will keep a pack of fuzzy peaches b) you will be able to keep any of the candy c) you will not be able to keep any of the candy

Answer 1

The probability that the person will be able to keep a pack of fuzzy peaches will be 0.25.

How to calculate the probability?

a. Based on the information given, the probability that the person will be able to keep a pack of fuzzy peaches will be:

= 10/(10 + 7 + 3) × 10/(10 + 7 + 3)

= 10/20 × 10/20

= 1/4

= 0.25

The probability that the person will be able to keep a pack of fuzzy peaches will be 0.25.

b. The probability to be able to keep any of the candy will be:

= (10/20 × 10/20) + (7/20 × 7/20) + (3/20 × 3/20)

= 100/400 + 49/400 + 9/400

= 158/400

= 0.395

The probability to be able to keep any of the candy will be 0.395.

c. The probability that you will not be able to keep any of the candy will be:

= (10/20 × 10/20) + (7/20 × 13/20) + (3/20 × 17/20)

= 100/400 + 91/400 + 51/400

= 242/400

= 0.605

The probability that you will not be able to keep any of the candy will be 0.605.

Learn more about probability on:

brainly.com/question/24756209

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