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Set S is the prime integers between 0 and 20. If three numbers are cho
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01 Nov 2021, 06:38
01 Nov 2021, 06:38
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Set S is the prime integers between 0 and 20. If three numbers are chosen randomly from set S, what is the probability that the sum of these three numbers is odd?
(A) 15/56
(B) 3/8
(C) 15/28
(D) 5/8
(E) 3/4
(A) 15/56
(B) 3/8
(C) 15/28
(D) 5/8
(E) 3/4
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Re: Set S is the prime integers between 0 and 20. If three numbers are cho
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01 Nov 2021, 06:50
01 Nov 2021, 06:50
1
Carcass wrote:
Set S is the prime integers between 0 and 20. If three numbers are chosen randomly from set S, what is the probability that the sum of these three numbers is odd?
(A) 15/56
(B) 3/8
(C) 15/28
(D) 5/8
(E) 3/4
(A) 15/56
(B) 3/8
(C) 15/28
(D) 5/8
(E) 3/4
Set S = {2, 3, 5, 7, 11, 13, 17, 19}
Important: Notice that 7 of the values are ODD, and 1 value is EVEN.
Also recognize that the sum of any 3 odd integers will always be ODD.
Conversely, the sum of 2 odd integers and 1 even integer will always be EVEN.
So, P(sum is odd) = P(all 3 selected numbers are odd)
= P(1st selected number is odd AND 2nd selected number is odd AND 3rd selected number is odd)
= P(1st selected number is odd) x P(2nd selected number is odd) x P(3rd selected number is odd)
= 7/8 x 6/7 x 5/6
= 5/8
Answer: D
Aside: P(1st selected number is odd) = 7/8, since 7 of the 8 numbers in the set are odd.
P(2nd selected number is odd) = 6/7, because once we select an odd number in the first draw, there are 7 numbers remaining, and 6 of them are odd.
etc....
Set S is the prime integers between 0 and 20. If three numbers are cho
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03 Nov 2021, 09:27
03 Nov 2021, 09:27
2
List of prime numbers from 0 to 20
\((2, 3, 5, 7, 11, 13, 17, 19)\)
There are 8 such numbers in total
Let us check in what cases adding three numbers gives an odd number
1. odd + odd + odd = [odd + odd] + odd = even + odd = odd
2. even + even + odd = [even + even] + odd = even + odd = odd - this wont apply since we only have 1 even number in our set which is \(2\)
Since all prime numbers except 2 are odd, we can choose from any set of three number which does not include \(2\).
Hence the probability will be \(\frac{^7C_3}{^8C_3} = \frac{7 * 6 * 5}{ 8 * 7 * 6} = \frac{5}{8}\)
\((2, 3, 5, 7, 11, 13, 17, 19)\)
There are 8 such numbers in total
Let us check in what cases adding three numbers gives an odd number
1. odd + odd + odd = [odd + odd] + odd = even + odd = odd
2. even + even + odd = [even + even] + odd = even + odd = odd - this wont apply since we only have 1 even number in our set which is \(2\)
Since all prime numbers except 2 are odd, we can choose from any set of three number which does not include \(2\).
Hence the probability will be \(\frac{^7C_3}{^8C_3} = \frac{7 * 6 * 5}{ 8 * 7 * 6} = \frac{5}{8}\)
