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Set A consists of the values 1, 2, and 3. Set B consists
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08 Oct 2018, 12:41
08 Oct 2018, 12:41
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25% (02:01) wrong based on 84 sessions
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Set A consists of the values 1, 2, and 3. Set B consists of the values 5, 6, and 7. One number is selected at random from each set. The two selected numbers are then added together.
The probability that the sum is even is how much greater than the probability that the sum is a prime number?
A. \(\frac{1}{9}\)
B. \(\frac{2}{9}\)
C. \(\frac{1}{3}\)
D. \(\frac{2}{5}\)
E. \(\frac{3}{5}\)
The probability that the sum is even is how much greater than the probability that the sum is a prime number?
A. \(\frac{1}{9}\)
B. \(\frac{2}{9}\)
C. \(\frac{1}{3}\)
D. \(\frac{2}{5}\)
E. \(\frac{3}{5}\)
Re: Set A consists of the values 1, 2, and 3. Set B consists
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14 Oct 2018, 22:16
14 Oct 2018, 22:16
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There are two different sets of numbers.
a sum of numbers is even if both the numbers are either odd or both are even
Hence possibilities for even numbers are {(1,5)(1,7)(3,5)(3,7)(2,6)}
total possibilities is 3*3 as there are 3 options for each set
probability for even = \(\frac{5}{9}\)
Prime numbers can be formed by two possibilities only {(1,6)(2,5)}
i.e \(\frac{2}{9}\)
\(\frac{5}{9}\) - \(\frac{2}{9}\) = \(\frac{3}{9}\) which reduces to \(\frac{1}{3}\)
a sum of numbers is even if both the numbers are either odd or both are even
Hence possibilities for even numbers are {(1,5)(1,7)(3,5)(3,7)(2,6)}
total possibilities is 3*3 as there are 3 options for each set
probability for even = \(\frac{5}{9}\)
Prime numbers can be formed by two possibilities only {(1,6)(2,5)}
i.e \(\frac{2}{9}\)
\(\frac{5}{9}\) - \(\frac{2}{9}\) = \(\frac{3}{9}\) which reduces to \(\frac{1}{3}\)
Re: Set A consists of the values 1, 2, and 3. Set B consists
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12 Dec 2018, 23:50
12 Dec 2018, 23:50
1
Hence possibilities for even numbers = 1 - ( All odd cases : Even + Odd )
Probability of Even + Odd
Case 1 => Even from Set A 1/3 ; Odd from Set B 2/3 => 1/3 * 2/3 = 2/9
Case 2 => Odd from Set A 2/3 ; Even from Set B 1/3 => 2/3 * 1/3 = 2/9
Total Probability of Even + Odd = 4/9 [ Case 1 + Case 2]
Possibilities for sum as even number = 1 - 4/9 = 5/9
Possibilities for sum as prime number = 2/9 => only 2 case (1,6) or (6,1)
Answer = Possibilities for sum as even number - Possibilities for sum as prime number => 5/9 - 2/9 => 3/9 => 1/3
Probability of Even + Odd
Case 1 => Even from Set A 1/3 ; Odd from Set B 2/3 => 1/3 * 2/3 = 2/9
Case 2 => Odd from Set A 2/3 ; Even from Set B 1/3 => 2/3 * 1/3 = 2/9
Total Probability of Even + Odd = 4/9 [ Case 1 + Case 2]
Possibilities for sum as even number = 1 - 4/9 = 5/9
Possibilities for sum as prime number = 2/9 => only 2 case (1,6) or (6,1)
Answer = Possibilities for sum as even number - Possibilities for sum as prime number => 5/9 - 2/9 => 3/9 => 1/3
