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Last year, Khalil came in $2$ nd place in his school's Spelling Bee. This year, he plans to win $1$ st place. He compiled a long list of words to study, $51\%$ of which have a Latin root. $\newline$ If Khalil randomly chooses a word to study from the list $2$ different times this month, what is the probability that $0$ of the words have a Latin root? $\newline$ Write your answer as a decimal rounded to the nearest thousandth. $\newline$ ____ $\newline$

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Find probabilities using the binomial distribution

Full solution

Q. Last year, Khalil came in

2

nd place in his school's Spelling Bee. This year, he plans to win

1

st place. He compiled a long list of words to study,

51\%

of which have a Latin root.

\newline

If Khalil randomly chooses a word to study from the list

2

different times this month, what is the probability that

0

of the words have a Latin root?

\newline

Write your answer as a decimal rounded to the nearest thousandth.

\newline

____

\newline

Find Probability Without Latin Root: First, we need to find the probability that one word does not have a Latin root. Since $51\%$ of the words have a Latin root, $100\% - 51\% = 49\%$ of the words do not have a Latin root.
Convert Percentage to Decimal: Now, we convert the percentage to a decimal to make calculations easier. So, $49\%$ as a decimal is $0.49$ .
Calculate Probability for Two Events: Next, we calculate the probability that Khalil chooses a word without a Latin root twice in a row. We multiply the probability of the first event by the probability of the second event, since the choices are independent. So, $0.49 \times 0.49$ .
Multiply Probabilities: Doing the multiplication, we get $0.49 \times 0.49 = 0.2401$ .
Round to Nearest Thousandth: Finally, we round the answer to the nearest thousandth as instructed. The rounded answer is $0.240$ .

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