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If x is to be chosen at random from the set {1,2,3,4} and y
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28 May 2020, 22:52
28 May 2020, 22:52
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If x is to be chosen at random from the set {1,2,3,4} and y is to be chosen at random from the set {5,6,7}, what is the probability that xy will be even?
A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 5/6
A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 5/6
Re: If x is to be chosen at random from the set {1,2,3,4} and y
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29 May 2020, 02:06
29 May 2020, 02:06
Expert Reply
1 - P(opposite event) = 1 - P(odd) = 1 - P(odd)*P(odd) = 1 - 2/4*2/3 = 8/12 = 2/3
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Re: If x is to be chosen at random from the set {1,2,3,4} and y
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25 Oct 2022, 10:02
25 Oct 2022, 10:02
Given that x is to be chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7} and we need to find what is the probability that xy will be even
We know that Total Probability is always equal to 1
=> P(Odd) + P(Even) = 1
=> P(Even) = 1 - P(Odd)
To get xy as odd we need to have all of them as odd.
=> P(xy = Odd) = P(x = odd) * P(y = odd) = \(\frac{2}{4}\) * \(\frac{2}{3}\)
(Because for x we have two choices (1,3) to get odd out of 4, for y we have two choices (5,7) to get odd out of 3)
= \(\frac{1}{3}\)
=> P(Even) = 1 - P(Odd) = 1 - \(\frac{1}{3}\) = \(\frac{3 - 1}{3}\) = \(\frac{2}{3}\)
So, Answer will be D.
Hope it helps!
We know that Total Probability is always equal to 1
=> P(Odd) + P(Even) = 1
=> P(Even) = 1 - P(Odd)
To get xy as odd we need to have all of them as odd.
=> P(xy = Odd) = P(x = odd) * P(y = odd) = \(\frac{2}{4}\) * \(\frac{2}{3}\)
(Because for x we have two choices (1,3) to get odd out of 4, for y we have two choices (5,7) to get odd out of 3)
= \(\frac{1}{3}\)
=> P(Even) = 1 - P(Odd) = 1 - \(\frac{1}{3}\) = \(\frac{3 - 1}{3}\) = \(\frac{2}{3}\)
So, Answer will be D.
Hope it helps!
