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by DBushkalov » Mon Apr 01, 2013 11:26 am
hi fellows,

here's one task for the community. I only want to know what i do wrong, but i will post a more in depth description of the prompt for those who want to solve it by themselves:

"If X is to be chosen at random fromthe set {1;2;3;4) and Y is to be chosen at random from the set {5;6;7}, what is the probability that X*Y will be even?"

A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 5/6

Here is how the GMAC guys calculate it:

there are 12 (4*3) possible products of X*Y where X is chosen from the first set and Y is chosen form the second set. The possible outcomes for 2 odd numbers are 4 (2*2) --> 4/12 is the probability that the product will be odd, thus 8/12 is the probability that the product will be EVEN.

Here is how I calculated it:

if we get an even number from the first set OR, i repeat, OR, an even number from the second set --> VIOLA ! we have an even number when X*Y. Lets now calculate the probability that we draw out an even number from the first and second sets:
1-st set: there are 2 even numbers from 4 ---> 2/4 = 1/2 is the probability of drawing an even nr. from the first set.

2-nd set: there is 1 even number ---> 1/3 is the probability to draw it out

From 1 and 2 follows that the total probability for an even number to be drawn is = 1/2 + 1/3, which yields 5/6.

5/6 is not 2/3 ,though. Can you please tell me what i do wrong ?

Thank you in advance :)
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Source: — Problem Solving |

by srcc25anu » Mon Apr 01, 2013 11:45 am
To get an even number we can have either even * odd OR even * even.
X = {1,2,3,4}
Y - {5,6,7}
EVEN from Y and ANY from X
Now if we draw any 1 number from Set X out of 4 numbers, and we select 6 from Y - we will get an EVEN number. those cases will be (1,6); (2,6); (3,6) ; (4,6)
so the prob will be 4/4 * 1/3 = 1/3
EVEN from X and ODD from Y
Next if we first draw an even number from X, we can select one of the odds (5,7). we are not counting 6 again as that would lead to double counting (of 2,6 and 4,6 - we have already considered these two in the above case).
so the cases will be (2,5) , (2,7), (4,5), (4,7)
hence total selections that will yield an EVEN number = 8
Total # selections = 4*3 =12
hence reqd probability = 8/12 = 2/3
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by hemant_rajput » Mon Apr 01, 2013 11:58 am
DBushkalov wrote:hi fellows,

here's one task for the community. I only want to know what i do wrong, but i will post a more in depth description of the prompt for those who want to solve it by themselves:

"If X is to be chosen at random fromthe set {1;2;3;4) and Y is to be chosen at random from the set {5;6;7}, what is the probability that X*Y will be even?"

A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 5/6

Here is how the GMAC guys calculate it:

there are 12 (4*3) possible products of X*Y where X is chosen from the first set and Y is chosen form the second set. The possible outcomes for 2 odd numbers are 4 (2*2) --> 4/12 is the probability that the product will be odd, thus 8/12 is the probability that the product will be EVEN.

Here is how I calculated it:

if we get an even number from the first set OR, i repeat, OR, an even number from the second set --> VIOLA ! we have an even number when X*Y. Lets now calculate the probability that we draw out an even number from the first and second sets:
1-st set: there are 2 even numbers from 4 ---> 2/4 = 1/2 is the probability of drawing an even nr. from the first set.

2-nd set: there is 1 even number ---> 1/3 is the probability to draw it out

From 1 and 2 follows that the total probability for an even number to be drawn is = 1/2 + 1/3, which yields 5/6.

5/6 is not 2/3 ,though. Can you please tell me what i do wrong ?

Thank you in advance :)
what you've done is you've listed getting even no. from the both the set but you've left the case when you've picked the even no. from one set and odd no. from other set.

so here is the correct approach.

(x even AND y odd) OR (x odd AND y even) OR (x even AND y even)

x even AND y odd - (2/4) * (2/3) = 1/3
x odd AND y even - (2/4) * (1/3) = 1/6
x even AND y even - (2/4) * (1/3) = 1/6

1/3 + 1/6 + 1/6 = 2/3

Hope this helps,

kudos,

Hemant
I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.
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