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If x and y are two positive integers and x + y = 5, then wha
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06 May 2019, 04:45
06 May 2019, 04:45
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If x and y are two positive integers and x + y = 5, then what is the probability that x equals 1?
(A) \(\frac{1}{2}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{4}\)
(D) \(\frac{1}{5}\)
(E) \(\frac{1}{6}\)
(A) \(\frac{1}{2}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{4}\)
(D) \(\frac{1}{5}\)
(E) \(\frac{1}{6}\)
Re: If x and y are two positive integers and x + y = 5, then wha
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13 Oct 2019, 02:06
13 Oct 2019, 02:06
The possible solutions are:
x y
1 4
2 3
3 2
4 1
Hence, x=1 is 1 case over 4 possible.
Answer C.
x y
1 4
2 3
3 2
4 1
Hence, x=1 is 1 case over 4 possible.
Answer C.
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If x and y are two positive integers and x + y = 5, then wha
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31 Aug 2022, 00:31
31 Aug 2022, 00:31
1
Theory: Probability of Occurrence of and event P(E) is given by
P(E) = Number of Favorable Outcomes / Total Number of Outcomes
Given that x and y are two positive integers and x + y = 5 and we need to find the probability that x equals 1
Let's start by writing all the possible outcomes for x+y = 5, given that both x and y are positive integers. Posssible outcomes are
x = 1, y = 4
x = 2, y = 3
x = 3, y = 2
x = 4, y = 1
So, Total Number of Outcomes = 4
Number of Favorable outcomes in which x = 1 is 1 (x = 1, y = 4)
=> Probability (x=1) = (Favorable outcomes in which x = 1) / Total Number of Outcomes = \(\frac{1}{4}\)
So, Answer will be C.
Hope it helps!
P(E) = Number of Favorable Outcomes / Total Number of Outcomes
Given that x and y are two positive integers and x + y = 5 and we need to find the probability that x equals 1
Let's start by writing all the possible outcomes for x+y = 5, given that both x and y are positive integers. Posssible outcomes are
x = 1, y = 4
x = 2, y = 3
x = 3, y = 2
x = 4, y = 1
So, Total Number of Outcomes = 4
Number of Favorable outcomes in which x = 1 is 1 (x = 1, y = 4)
=> Probability (x=1) = (Favorable outcomes in which x = 1) / Total Number of Outcomes = \(\frac{1}{4}\)
So, Answer will be C.
Hope it helps!
