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18 Dec 2020, 03:30
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
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Question Stats:
71% (01:35) correct
29%
(01:45)
wrong
based on 94
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A box contains five blocks numbered 1, 2, 3, 4, and 5. John picks a block and replaces it. Lisa then picks a block. What is the probability that the sum of the numbers they picked is even?
(A) 9/25
(B) 2/5
(C) 1/2
(D) 13/25
(E) 3/5
(A) 9/25
(B) 2/5
(C) 1/2
(D) 13/25
(E) 3/5
18 Dec 2020, 04:09
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Both events are interdependent. So, we need to multiple them.
Case1: when we have 2 or 4 available for selection, P(E1) = {2C1/5C1}*{2C1/5C1}= 4/25
Case2: when we have 1, 3 or 5 available for selection, P(PE2)= {3C1/5C1}*{3C1/5C1}= 9/25
So, P(E) = 4/25 + 9/25= 13/25.
Ans. D.
Case1: when we have 2 or 4 available for selection, P(E1) = {2C1/5C1}*{2C1/5C1}= 4/25
Case2: when we have 1, 3 or 5 available for selection, P(PE2)= {3C1/5C1}*{3C1/5C1}= 9/25
So, P(E) = 4/25 + 9/25= 13/25.
Ans. D.
18 Dec 2020, 05:05
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Even + Even = Even (Both picks even numbered block)
Odd + Odd = Even (Both picks odd numbered block)
P(Even) = \(\frac{2}{5}\) and P(Odd) = \(\frac{3}{5}\)
Both picks Even: \(\frac{2}{5}\) * \(\frac{2}{5}\) (Independent case)= \(\frac{4}{25}\)
Both picks Odd: \(\frac{3}{5}\) * \(\frac{3}{5}\) (Independent case) = \(\frac{9}{25}\)
Total: \(\frac{4}{25}\) + \(\frac{9}{25}\) = \(\frac{13}{25}\)
Answer D
Odd + Odd = Even (Both picks odd numbered block)
P(Even) = \(\frac{2}{5}\) and P(Odd) = \(\frac{3}{5}\)
Both picks Even: \(\frac{2}{5}\) * \(\frac{2}{5}\) (Independent case)= \(\frac{4}{25}\)
Both picks Odd: \(\frac{3}{5}\) * \(\frac{3}{5}\) (Independent case) = \(\frac{9}{25}\)
Total: \(\frac{4}{25}\) + \(\frac{9}{25}\) = \(\frac{13}{25}\)
Answer D
