234k views
3 votes
Kai wrote the characters to the phrase 567 GNATS on separate chips and put them in a container. She selected one tile at random.

Find P (odd number or a consonant).

(1)/(4) (5)/(8) (3)/(4) (1)/(2)

Find the complement of P (vowel).

(5)/(8) (1)/(8) (7)/(8) (3)/(4)

If a tile is drawn from the hat 1,232 times, how many times would you expect the tile to have any of the characters from the phrase 6 TANS?
924 times
1078 times
154 times
770 times

1 Answer

3 votes

Part 1

Answer: 3/4

------------------

Step-by-step explanation:

set A = set of things we want to pull out

set A = set of odd numbers or consonants

set A = {5, 7, G, N, T, S}

There are 6 items in set A. Let p = 6.

set B = set of all possible outcomes

set B = {5, 6, 7, G, N, A, T, S}

There are three numbers and five letters, so 3+5 = 8 items total in set B. Let q = 8.

Divide the values of p and q to get the final result

p/q = 6/8 = 3/4

================================================

Part 2

Answer: 7/8

------------------

Step-by-step explanation:

There is one vowel (the letter "A") out of 8 items total. So there are 8-1 = 7 items that aren't a vowel. The complement of P(vowel) is basically P(not vowel) and that's why the answer is 7/8

================================================

Part 3

Answer: 924

------------------

Step-by-step explanation:

The phrase "6 TANS" has five unique items, either a number or the letters mentioned. This is out of 8 items total. So 6/8 = 3/4 is the probability of picking any of those items. If we do 1232 trials, then we expect (3/4)*1232 = 924 occurrences where that number 6 or those letters show up. This value is an estimate.

User Mamuz
by
8.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories