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16 Sep 2014, 00:36
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\(A = \{2, 3, 4, 4, 4\}\)
\(B = \{0, 1, 2\}\)
If a number is taken from list \(A\) at random and another number is taken from list \(B\) at random, what is the probability that the sum of these numbers is a prime number?
A. \(\frac{1}{15}\)
B. \(\frac{2}{15}\)
C. \(\frac{5}{15}\)
D. \(\frac{7}{15}\)
E. \(\frac{9}{15}\)
\(B = \{0, 1, 2\}\)
If a number is taken from list \(A\) at random and another number is taken from list \(B\) at random, what is the probability that the sum of these numbers is a prime number?
A. \(\frac{1}{15}\)
B. \(\frac{2}{15}\)
C. \(\frac{5}{15}\)
D. \(\frac{7}{15}\)
E. \(\frac{9}{15}\)
16 Sep 2014, 00:36
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Official Solution:
\(A = \{2, 3, 4, 4, 4\}\)
\(B = \{0, 1, 2\}\)
If a number is taken from list \(A\) at random and another number is taken from list \(B\) at random, what is the probability that the sum of these numbers is a prime number?
A. \(\frac{1}{15}\)
B. \(\frac{2}{15}\)
C. \(\frac{5}{15}\)
D. \(\frac{7}{15}\)
E. \(\frac{9}{15}\)
The total number of possible outcomes is \(5*3 = 15\).
Now, since list \(B\) is smaller, let's work with it:
• 0 can be added to 2 or to 3 to get a prime: \(1+1=2\) cases;
• 1 can be added to 2 or to either of three 4's to get a prime: \(1+3=4\) cases;
• 2 can be added only to 3 to get a prime: 1 case.
Hence, the total number of favorable outcomes is \(2+4+1=7\).
Therefore, the probability is \(P=\frac{\text{favorable} }{\text{total} }=\frac{7}{15}\).
Answer: D
\(A = \{2, 3, 4, 4, 4\}\)
\(B = \{0, 1, 2\}\)
If a number is taken from list \(A\) at random and another number is taken from list \(B\) at random, what is the probability that the sum of these numbers is a prime number?
A. \(\frac{1}{15}\)
B. \(\frac{2}{15}\)
C. \(\frac{5}{15}\)
D. \(\frac{7}{15}\)
E. \(\frac{9}{15}\)
The total number of possible outcomes is \(5*3 = 15\).
Now, since list \(B\) is smaller, let's work with it:
• 0 can be added to 2 or to 3 to get a prime: \(1+1=2\) cases;
• 1 can be added to 2 or to either of three 4's to get a prime: \(1+3=4\) cases;
• 2 can be added only to 3 to get a prime: 1 case.
Hence, the total number of favorable outcomes is \(2+4+1=7\).
Therefore, the probability is \(P=\frac{\text{favorable} }{\text{total} }=\frac{7}{15}\).
Answer: D
12 Dec 2019, 07:58
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Here's a video explanation. Enjoy!
