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Suppose a, b, c, d, e are selected randomly from the set
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22 Nov 2020, 18:45
22 Nov 2020, 18:45
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Suppose a, b, c, d, e are selected randomly from the set {1, 2, 3, 4, 5} and they can repeat. Find the probability that a*b*c*d+e is odd.
A.\(\frac{12}{25}\)
B.\(\frac{27}{125}\)
C.\(\frac{243}{3125}\)
D.\(\frac{1632}{3125}\)
E.\(\frac{1794}{3125}\)
A.\(\frac{12}{25}\)
B.\(\frac{27}{125}\)
C.\(\frac{243}{3125}\)
D.\(\frac{1632}{3125}\)
E.\(\frac{1794}{3125}\)
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Re: Suppose a, b, c, d, e are selected randomly from the set
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23 Nov 2020, 12:48
23 Nov 2020, 12:48
1
lazyashell wrote:
Suppose a, b, c, d, e are selected randomly from the set {1, 2, 3, 4, 5} and they can repeat. Find the probability that a*b*c*d+e is odd.
A.\(\frac{12}{25}\)
B.\(\frac{27}{125}\)
C.\(\frac{243}{3125}\)
D.\(\frac{1632}{3125}\)
E.\(\frac{1794}{3125}\)
A.\(\frac{12}{25}\)
B.\(\frac{27}{125}\)
C.\(\frac{243}{3125}\)
D.\(\frac{1632}{3125}\)
E.\(\frac{1794}{3125}\)
P(abcd + e is odd) = P(abcd is odd AND e is even OR abcd is even AND e is odd)
= P(abcd is odd AND e is even) + P(abcd is even AND e is odd)
Let's examine each probability separately...
P(abcd is odd AND e is even) = P(abcd is odd) x P(e is even)
---Aside----------------------
In order for abcd to be odd, all four values (a, b, c, and d) must be odd.
P(a selected number is odd = (3/5)
So, P(a, b, c, d are all odd) = (3/5)(3/5)(3/5)(3/5) = 81/625
In other words, P(abcd is odd) = 81/625
-----------------------------
So, P(abcd is odd) x P(e is even) = 81/625 x 2/5 = 162/3125
P(abcd is even AND e is odd) = P(abcd is even) x P(e is odd)
---Aside----------------------
We already learned that P(abcd is odd) = 81/625
This means P(abcd is NOT odd) = 1 - 81/625 = 544/625
In other words, P(abcd is EVEN) = 544/625
-----------------------------
So, P(abcd is even) x P(e is odd) = 544/625 x 3/5 = 1632/3125
We get: P(abcd is odd AND e is even) + P(abcd is even AND e is odd) = 162/3125 + 1632/3125
= 1794/3125
Answer: E
Cheers,
Brent
Re: Suppose a, b, c, d, e are selected randomly from the set
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24 Nov 2020, 00:03
24 Nov 2020, 00:03
GreenlightTestPrep wrote:
lazyashell wrote:
Suppose a, b, c, d, e are selected randomly from the set {1, 2, 3, 4, 5} and they can repeat. Find the probability that a*b*c*d+e is odd.
A.\(\frac{12}{25}\)
B.\(\frac{27}{125}\)
C.\(\frac{243}{3125}\)
D.\(\frac{1632}{3125}\)
E.\(\frac{1794}{3125}\)
A.\(\frac{12}{25}\)
B.\(\frac{27}{125}\)
C.\(\frac{243}{3125}\)
D.\(\frac{1632}{3125}\)
E.\(\frac{1794}{3125}\)
P(abcd + e is odd) = P(abcd is odd AND e is even OR abcd is even AND e is odd)
= P(abcd is odd AND e is even) + P(abcd is even AND e is odd)
Let's examine each probability separately...
P(abcd is odd AND e is even) = P(abcd is odd) x P(e is even)
---Aside----------------------
In order for abcd to be odd, all four values (a, b, c, and d) must be odd.
P(a selected number is odd = (3/5)
So, P(a, b, c, d are all odd) = (3/5)(3/5)(3/5)(3/5) = 81/625
In other words, P(abcd is odd) = 81/625
-----------------------------
So, P(abcd is odd) x P(e is even) = 81/625 x 2/5 = 162/3125
P(abcd is even AND e is odd) = P(abcd is even) x P(e is odd)
---Aside----------------------
We already learned that P(abcd is odd) = 81/625
This means P(abcd is NOT odd) = 1 - 81/625 = 544/625
In other words, P(abcd is EVEN) = 544/625
-----------------------------
So, P(abcd is even) x P(e is odd) = 544/625 x 3/5 = 1632/3125
We get: P(abcd is odd AND e is even) + P(abcd is even AND e is odd) = 162/3125 + 1632/3125
= 1794/3125
Answer: E
Cheers,
Brent
is it a standard question? we can consider other arrangements with different answers. like O*O*E*E+O=O or O*O*O*O+E=O or E*E*E*E+O=O and.... so it depends on you which arrangement you want to choose with different answers. I don't know which part I missed that think so
