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02 Aug 2023, 01:30
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E
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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.\(\frac{1794}{3125}\)
02 Aug 2023, 22:40
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Bunuel
Question : Probability of (a*b*c*d+e = odd)
- Even Terms = {2, 4}
- Odd Terms = {1, 3, 5}
As the addition of two terms equals odd, the terms can be -
Case 1 : a*b*c*d = even AND e = odd
Even + Odd = Odd
Case 2 : a*b*c*d = odd AND e = even
Odd + Even = Odd
Case 1 : a*b*c*d = even AND e = odd
a*b*c*d = even
Inference: Among a, b, c, d, and e one term must be even; put otherwise none of the terms should be odd.
Number of ways = Total Number of ways - (All the terms are odd)
Without any restriction, each term can be selected in 5 ways
Number of ways the terms can be selected such that a*b*c*d equals even = \(5^4 - 3^4\)
e = odd
'e' can be selected in three ways.
Number of ways to perform the selection = \((5^4 - 3^4) * 3 = 1632\)
Case 2 : a*b*c*d = odd AND e = even
a*b*c*d = odd
Because the product of the terms is odd, each of the terms should be odd. a, b, c, and d can be selected in \(3\) ways.
Number of ways = \(3^4\)
e = even
'e' can be selected in two ways.
Number of ways to perform the selection = \((3^4) * 2 = 162\)
Required Probaility
Total Favourbale cases = \(162 + 1632 = 1794\)
Total Number of ways the selections can be made = \(5^5 = 3125\)
Required Probability = \(\frac{1794 }{ 3125}\)
Option E
General Discussion
04 Aug 2023, 05:16
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(abcd)+e=Odd
2 cases are possible
Case 1
abcd=Even and e=Odd
Within abcd
One number is Even=4C1*2*3*3*3=216
Two numbers are Even=4C2*2*2*3*3=216
Three numbers are Even=4C3*2*2*2*3=96
All four numbers are Even=4C4*2*2*2*2=16
Total cases abcd is even=544
With e
e has 3 possibilities
Total cases for Case 1=3*544=1632
Case 2
abcd=Odd and e=Even
Within abcd
Total cases=abcd is even+abcd is odd
Total cases abcd is odd=625-544=81
With e
e has 2 possibilities
Total cases for Case 2=81*2=162
Hence total possible cases with conditions = 1632+162= 1794
Total possible cases without conditions= 5*5*5*5*5=3125
Probability=1794/3125
Answer Option E.
2 cases are possible
Case 1
abcd=Even and e=Odd
Within abcd
One number is Even=4C1*2*3*3*3=216
Two numbers are Even=4C2*2*2*3*3=216
Three numbers are Even=4C3*2*2*2*3=96
All four numbers are Even=4C4*2*2*2*2=16
Total cases abcd is even=544
With e
e has 3 possibilities
Total cases for Case 1=3*544=1632
Case 2
abcd=Odd and e=Even
Within abcd
Total cases=abcd is even+abcd is odd
Total cases abcd is odd=625-544=81
With e
e has 2 possibilities
Total cases for Case 2=81*2=162
Hence total possible cases with conditions = 1632+162= 1794
Total possible cases without conditions= 5*5*5*5*5=3125
Probability=1794/3125
Answer Option E.
