Question

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

Answer 1

Part 1

Answer: 3/4

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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

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Part 2

Answer: 7/8

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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

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Part 3

Answer: 924

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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.