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When x, y are selected randomly from set {0,1,2,3,4,5,6,7,8,

Expert replies
by Max@Math Revolution » Tue Jan 19, 2016 6:07 pm
When x, y are selected randomly from set {0,1,2,3,4,5,6,7,8,9}, what is the probability that the product of x and y selected is even?

A. 1/4
B. 1/2
C. 3/4
D. 4/5
E. 9/10


A solution will be posted in two days.
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Source: — Problem Solving |

by GMATinsight » Wed Jan 20, 2016 6:08 am
Max@Math Revolution wrote:When x, y are selected randomly from set {0,1,2,3,4,5,6,7,8,9}, what is the probability that the product of x and y selected is even?

A. 1/4
B. 1/2
C. 3/4
D. 4/5
E. 9/10


A solution will be posted in two days.
Method-1:

Probability of Product x*y being odd = Prob of x being odd * Prob of y being odd
i.e. Probability of Product x*y being odd = (5/10)*(5/10) = 1/4
i.e.Probability of product x*y being even = 1- Probability of Product x*y being odd
i.e.Probability of product x*y being even = 1- (1/4) = 3/4

Answer: Option C
Last edited by GMATinsight on Wed Jan 20, 2016 6:13 am, edited 1 time in total.
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by GMATinsight » Wed Jan 20, 2016 6:12 am
Max@Math Revolution wrote:When x, y are selected randomly from set {0,1,2,3,4,5,6,7,8,9}, what is the probability that the product of x and y selected is even?

A. 1/4
B. 1/2
C. 3/4
D. 4/5
E. 9/10


A solution will be posted in two days.
Method-2:

Probability of Product x*y being Even =
Case 1: Prob of x being even * Prob of y being even = (5/10)*(5/10) = 1/4
Case 2: Prob of x being even * Prob of y being Odd = (5/10)*(5/10) = 1/4
Case 3: Prob of x being Odd * Prob of y being even = (5/10)*(5/10) = 1/4

i.e. Probability of Product x*y being Even = (1/4) + (1/4) + (1/4) = 3/4

Answer: Option C
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by Max@Math Revolution » Thu Jan 21, 2016 8:45 pm
When x, y are selected randomly from set {0,1,2,3,4,5,6,7,8,9}, what is the probability that the product of x and y selected is even?

A. 1/4 B. 1/2 C. 3/4 D. 4/5 E. 9/10

-> Probability=the number of cases of a certain event/the number of the whole cases.
The probability of multiplication becoming a even number=1-the probability of multiplication becoming an odd number=1-(5C1*5C1/10C1*10C1)=1-(5*5/10*10)=1-1/4=3/4. Therefore, the answer is C.
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