Question
A random selection is made from a collection comprising all integers between 10 and 40 inclusive. Determine the probabilities for the following scenarios: a) π(πΈπ£ππ | πΊππππ‘ππ π‘βππ 30) b) π(πΊππππ‘ππ π‘βππ 30 | πΈπ£ππ) c) π(πππππ | π΅ππ‘π€πππ 20 πππ 30 )
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Answer to a math question A random selection is made from a collection comprising all integers between 10 and 40 inclusive. Determine the probabilities for the following scenarios: a) π(πΈπ£ππ | πΊππππ‘ππ π‘βππ 30) b) π(πΊππππ‘ππ π‘βππ 30 | πΈπ£ππ) c) π(πππππ | π΅ππ‘π€πππ 20 πππ 30 )
Clarabelle
4.794 Answers
a) To find π(πΈπ£ππ | πΊππππ‘ππ π‘βππ 30), we first find the total number of even numbers between 10 and 40 (inclusive). The even numbers between 10 and 40 are {10, 12, 14, ..., 38, 40}, which is a total of 16 numbers.
The numbers greater than 30 are {31, 32, ..., 40}, which is a total of 10 numbers. Out of these, the even numbers are {32, 34, 36, 38, 40}, which is a total of 5 numbers.
Therefore, π(πΈπ£ππ | πΊππππ‘ππ π‘βππ 30) =\frac{5}{10} \frac{1}{2}
b) To find π(πΊππππ‘ππ π‘βππ 30 | πΈπ£ππ), we first find the total number of even numbers between 10 and 40 (inclusive), which is 16.
The number of even numbers greater than 30 is 5 as found in part (a).
Therefore, π(πΊππππ‘ππ π‘βππ 30 | πΈπ£ππ) =\frac{5}{16}
c) To find π(πππππ | π΅ππ‘π€πππ 20 πππ 30), we need to find the prime numbers between 20 and 30. The prime numbers between 20 and 30 are {23, 29}, which is a total of 2 numbers.
The numbers between 20 and 30 (inclusive) are {20, 21, ..., 30}, which is a total of 11 numbers.
Therefore, π(πππππ | π΅ππ‘π€πππ 20 πππ 30) =\frac{2}{11}
\textbf{Answer:}
a)\frac{1}{2}
b)\frac{5}{16}
c)\frac{2}{11}
The numbers greater than 30 are {31, 32, ..., 40}, which is a total of 10 numbers. Out of these, the even numbers are {32, 34, 36, 38, 40}, which is a total of 5 numbers.
Therefore, π(πΈπ£ππ | πΊππππ‘ππ π‘βππ 30) =
b) To find π(πΊππππ‘ππ π‘βππ 30 | πΈπ£ππ), we first find the total number of even numbers between 10 and 40 (inclusive), which is 16.
The number of even numbers greater than 30 is 5 as found in part (a).
Therefore, π(πΊππππ‘ππ π‘βππ 30 | πΈπ£ππ) =
c) To find π(πππππ | π΅ππ‘π€πππ 20 πππ 30), we need to find the prime numbers between 20 and 30. The prime numbers between 20 and 30 are {23, 29}, which is a total of 2 numbers.
The numbers between 20 and 30 (inclusive) are {20, 21, ..., 30}, which is a total of 11 numbers.
Therefore, π(πππππ | π΅ππ‘π€πππ 20 πππ 30) =
\textbf{Answer:}
a)
b)
c)
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