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Two integers between 1 and 100, inclusive, each randomly and independe
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24 Apr 2015, 03:03
24 Apr 2015, 03:03
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Two integers between 1 and 100, inclusive, each randomly and independently chosen, are either added or multiplied, with an equal chance of either operation. What is the probability that the result is even?
(A) 1/3
(B) 1/2
(C) 5/8
(D) 2/3
(E) 7/8
(A) 1/3
(B) 1/2
(C) 5/8
(D) 2/3
(E) 7/8
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Re: Two integers between 1 and 100, inclusive, each randomly and independe
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24 Apr 2015, 11:57
24 Apr 2015, 11:57
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Bunuel
Two integers between 1 and 100, inclusive, each randomly and independently chosen, are either added or multiplied, with an equal chance of either operation. What is the probability that the result is even?
(A) 1/3
(B) 1/2
(C) 5/8
(D) 2/3
(E) 7/8
Kudos for a correct solution.
(A) 1/3
(B) 1/2
(C) 5/8
(D) 2/3
(E) 7/8
Kudos for a correct solution.
Probability of getting even = 1 - Probability of getting Odd
Getting Odd outcome = Odd*Odd or Odd + Even or Even + Odd
Thus, Probability of picking Odd, Odd and multiplication = \(1/2*1/2*1/2 = 1/8\)
Probability of picking Odd, Even and addition = \(1/2*1/2*1/2 = 1/8\)
Probability of picking Even, Odd and addition = \(1/2*1/2*1/2 = 1/8\)
Thus, Probability of getting an Odd number = \(1/8 + 1/8 + 1/8 = 3/8\)
And Probability of getting an Even number = \((1 - 3/8) = 5/8\)
Hence, answer is C
Re: Two integers between 1 and 100, inclusive, each randomly and independe
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25 Apr 2015, 06:02
25 Apr 2015, 06:02
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Total number of integers = 100
Number of Even integers = 50
Number of Odd integers = 50
Probability of selecting an odd integer = \(\frac{50}{100}\) = \(\frac{1}{2}\)
Probability of selecting an even integer = \(\frac{50}{100}\) = \(\frac{1}{2}\)
From number properties, we know that for addition of two integers to be even => Odd + Odd = Even or Even + Even = Even
Using the above cases, the probability that the addition of two integers is even (Pa)= \([\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{1}{2}]\) = \(\frac{1}{4} + \frac{1}{4}\) = \(\frac{1}{2}\)
Similarly , from number properties, we know that for multiplication of two integers to be even => Even * Odd = Even or Odd * Even = Even or Even * Even = Even
Using the above cases, the probability that the multiplication of two integers is even (Pm)= \([\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{1}{2}]\) = \(\frac{1}{4} + \frac{1}{4} + \frac{1}{4}\) = \(\frac{3}{4}\)
From the question stem , we know that both addition and multiplication have an equal chance i.e. \(\frac{1}{2}\) * Pa and \(\frac{1}{2}\) * Pm
Therefore, the total Probability for the result to be even = \(\frac{1}{2}\) * Pa + \(\frac{1}{2}\) * Pm = \([\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{3}{4}]\) = \(\frac{1}{4} + \frac{3}{8}\) = \(\frac{5}{8}\)
Answer is (C)
Number of Even integers = 50
Number of Odd integers = 50
Probability of selecting an odd integer = \(\frac{50}{100}\) = \(\frac{1}{2}\)
Probability of selecting an even integer = \(\frac{50}{100}\) = \(\frac{1}{2}\)
From number properties, we know that for addition of two integers to be even => Odd + Odd = Even or Even + Even = Even
Using the above cases, the probability that the addition of two integers is even (Pa)= \([\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{1}{2}]\) = \(\frac{1}{4} + \frac{1}{4}\) = \(\frac{1}{2}\)
Similarly , from number properties, we know that for multiplication of two integers to be even => Even * Odd = Even or Odd * Even = Even or Even * Even = Even
Using the above cases, the probability that the multiplication of two integers is even (Pm)= \([\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{1}{2}]\) = \(\frac{1}{4} + \frac{1}{4} + \frac{1}{4}\) = \(\frac{3}{4}\)
From the question stem , we know that both addition and multiplication have an equal chance i.e. \(\frac{1}{2}\) * Pa and \(\frac{1}{2}\) * Pm
Therefore, the total Probability for the result to be even = \(\frac{1}{2}\) * Pa + \(\frac{1}{2}\) * Pm = \([\frac{1}{2} * \frac{1}{2}] + [\frac{1}{2} * \frac{3}{4}]\) = \(\frac{1}{4} + \frac{3}{8}\) = \(\frac{5}{8}\)
Answer is (C)
