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Nikolay [14]
3 years ago
9

The table shows the snack preferences of 50 shoppers at the mall. what is the probability that one shopper, selected at random f

rom the 50 surveyed, preferred the potato chips or pretzels?
A.1/5
B.13/25
C.1/10
D.18/25

Mathematics
2 answers:
Ipatiy [6.2K]3 years ago
6 0
1/5 is the answer homie

Gala2k [10]3 years ago
3 0

Answer: Option 'B' is correct.

Step-by-step explanation:

Since we have given that

Number of people who prefer chips = 12

Number of people who prefer pretzels = 14

Total number of people = 50

So, Probability that one preferred the potato chips or pretzels is given by

\dfrac{12}{50}+\dfrac{14}{50}\\\\=\dfrac{26}{50}\\\\=\dfrac{13}{25}

Probability that one preferred the potato chips or pretzels is \dfrac{13}{25}

Hence, Option 'B' is correct.

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Each letter of the word "supercalifragilisticexpialidocious" is placed into a bag and drawn at 3 times, replacing the letter aft
Makovka662 [10]

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P(X ≥ 1) = 0.50

Step-by-step explanation:

Given that:

The word "supercalifragilisticexpialidocious" has 34 letters in which 'i' appears 7 times in the word.

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Using Binomial distribution to determine the probability; we have:

P(X = x)  = ^nC_x  \ \beta^x   \  (1 - \beta)^{n-x}

where;

x = 0,1,2,...n    and    0  <  β   <   1

and x represents the  number of successes.

However; since the letter is drawn thrice; the probability that the letter "i" is drawn at least once can be computed as:

P(X ≥ 1) = 1 - P(X< 1)

P(X ≥ 1) = 1 - P(X =0)

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Answer:

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Step-by-step explanation:

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2    2

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