Q. Three boys ages 4, 6 and 7 respectively. Three girls ages 5,8 and 9 respectively. If two of the boys and two of the girls are randomly selected, and the sum of the selected children's age is z, what is the difference between the probability that z is even and the probability that z id odd?
A. 1/9
B. 1/6
C. 2/9
D. 1/4
E. 1/2
Please help me in understanding why answer is 1/9.
Probabilty Problem
This topic has expert replies
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Total ways to choose the 4 children:ras-j wrote:Q. Three boys ages 4, 6 and 7 respectively. Three girls ages 5,8 and 9 respectively. If two of the boys and two of the girls are randomly selected, and the sum of the selected children's age is z, what is the difference between the probability that z is even and the probability that z id odd?
A. 1/9
B. 1/6
C. 2/9
D. 1/4
E. 1/2
Number of pairs of boys that can be formed from 3 options = 3C2 = 3.
Number of pairs of girls that can be formed from 3 options = 3C2 = 3.
To combine these options, we multiply:
3*3 = 9.
Ways to get an ODD sum:
For the sum of 4 integers to be odd, either ONE of the integers or THREE of the integers must be odd.
Case 1: ONE integer is odd
(4,6) from the boys and (5,8) from the girls, yielding 4+6+5+8 = 23.
(4,6) from the boys and (8,9) from the girls, yielding 4+6+8+9 = 27.
Case 2: THREE integers are odd
(4,7) from the boys and (5,9) from the girls, yielding 4+7+5+9 = 25.
(6,7) from the boys and (5,9) from the girls, yielding 6+7+5+9 = 27.
Total ways to get an ODD sum = 4.
Thus, P(odd sum) = (total ways to get an odd sum)/(total ways to choose the 4 children) = 4/9.
Since P(odd sum) = 4/9, P(even sum) = 1 - 4/9 = 5/9.
Thus:
P(even sum) - P(odd sum) = 5/9 - 4/9 = 1/9.
.
The correct answer is A.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Here's a different approach.ras-j wrote:Q. Three boys ages 4, 6 and 7 respectively. Three girls ages 5,8 and 9 respectively. If two of the boys and two of the girls are randomly selected, and the sum of the selected children's age is z, what is the difference between the probability that z is even and the probability that z id odd?
A. 1/9
B. 1/6
C. 2/9
D. 1/4
E. 1/2
Notice that the sum of all 6 ages is 39. Rather than choose 2 boys and 2 girls, let's just remove one child from each group. This will still result in 2 boys and 2 girls.
P(sum of 4 ages is odd)
Since the sum of all 6 ages is odd, we can get an odd sum of 4 children if we remove two odds or remove two evens.
So, P(sum of 4 ages is odd) = P(odd boy and odd girl OR even boy and even girl)
= P(odd boy and odd girl) + P(even boy and even girl)
= (1/3)(2/3) + (2/3)(1/3)
= 2/9 + 2/9
= 4/9
P(sum of 4 ages is even)
P(sum of 4 ages is even) = 1 - P(sum of 4 ages is not even)
= 1 - P(sum of 4 ages is odd)
= 1 - 4/9
= 5/9
So, P(z is even) - P(z is odd) = 5/9 - 4/9 = 1/9 = A
Cheers,
Brent
My answer is 1/9
Approach is as follows
1. Total number of ways = 3c2*3c2 = 9
2. No. of pairs of boys such that sum is = 2 O, 1 E = O
3. No. of pairs of Girls = 2 o, 1 e
Case one - For Z to Even
Number of ways - E e, Oo1, Oo2, oO1, oO2 = 5
Probability - 5/9
Case two - For z to be ODD
Number of ways - Eo1, Eo2, eO1, eO2 = 4
Probbility - 4/9
Difference = 1/9
Hope i am using correct approach..
Approach is as follows
1. Total number of ways = 3c2*3c2 = 9
2. No. of pairs of boys such that sum is = 2 O, 1 E = O
3. No. of pairs of Girls = 2 o, 1 e
Case one - For Z to Even
Number of ways - E e, Oo1, Oo2, oO1, oO2 = 5
Probability - 5/9
Case two - For z to be ODD
Number of ways - Eo1, Eo2, eO1, eO2 = 4
Probbility - 4/9
Difference = 1/9
Hope i am using correct approach..
- Anaira Mitch
- Master | Next Rank: 500 Posts
- Posts: 235
- Joined: Wed Oct 26, 2016 9:21 pm
- Thanked: 3 times
- Followed by:5 members
-
- GMAT Instructor
- Posts: 2630
- Joined: Wed Sep 12, 2012 3:32 pm
- Location: East Bay all the way
- Thanked: 625 times
- Followed by:119 members
- GMAT Score:780
I didn't answer, but I have to chime in and speak for the group: we love the appreciation!Anaira Mitch wrote:Excellent Solution experts as always. Thank You!!
-
- GMAT Instructor
- Posts: 2630
- Joined: Wed Sep 12, 2012 3:32 pm
- Location: East Bay all the way
- Thanked: 625 times
- Followed by:119 members
- GMAT Score:780
But now that I've found the thread, I can't resist a solution of my own.
Once we know that they're distinct, the actual ages of each child aren't relevant, only their parity: whether they're even or odd. With that in mind, we've got two sets:
Boys: {E, E, O}
Girls: {O, E, O}
Choosing four of these, we'll get an odd sum with one odd or three odds, and an even sum with two odds. (Zero odds and four odds aren't possible, of course.)
Since the even sum is easier (only one calculation), let's do that, and find the probability of exactly two odds.
We'll pick two odds in one of two ways:
1) We take the odd boy and exactly one of the odd girls;
2) We don't take the odd boy, but we take both odd girls
Case #1:
Odd Boy * (Exactly One Odd Girl) =
(1 - Both Even Boys) * (1 - Both Odd Girls) =
(1 - 1/3) * (1 - 1/3) = 4/9
Case #2:
Both Even Boys * Both Odd Girls =
(2/3 * 1/2) * (2/3 * 1/2) = 1/9
So our probability of exactly two odds is the sum of these cases, 4/9 + 1/9 = 5/9.
The probability of NOT exactly two odds is 1 - (exactly two odds) = 1 - 5/9 = 4/9.
So the difference between the two is 5/9 - 4/9, or 1/9, and we're done!
Once we know that they're distinct, the actual ages of each child aren't relevant, only their parity: whether they're even or odd. With that in mind, we've got two sets:
Boys: {E, E, O}
Girls: {O, E, O}
Choosing four of these, we'll get an odd sum with one odd or three odds, and an even sum with two odds. (Zero odds and four odds aren't possible, of course.)
Since the even sum is easier (only one calculation), let's do that, and find the probability of exactly two odds.
We'll pick two odds in one of two ways:
1) We take the odd boy and exactly one of the odd girls;
2) We don't take the odd boy, but we take both odd girls
Case #1:
Odd Boy * (Exactly One Odd Girl) =
(1 - Both Even Boys) * (1 - Both Odd Girls) =
(1 - 1/3) * (1 - 1/3) = 4/9
Case #2:
Both Even Boys * Both Odd Girls =
(2/3 * 1/2) * (2/3 * 1/2) = 1/9
So our probability of exactly two odds is the sum of these cases, 4/9 + 1/9 = 5/9.
The probability of NOT exactly two odds is 1 - (exactly two odds) = 1 - 5/9 = 4/9.
So the difference between the two is 5/9 - 4/9, or 1/9, and we're done!
-
- GMAT Instructor
- Posts: 2630
- Joined: Wed Sep 12, 2012 3:32 pm
- Location: East Bay all the way
- Thanked: 625 times
- Followed by:119 members
- GMAT Score:780
Counting the number of cases is often a nice way to solve, so long as you know that the probability of each case is the same. (Here, any group of four children is equally likely, so counting isn't bad.)ashg84 wrote:My answer is 1/9
Approach is as follows
1. Total number of ways = 3c2*3c2 = 9
2. No. of pairs of boys such that sum is = 2 O, 1 E = O
3. No. of pairs of Girls = 2 o, 1 e
Case one - For Z to Even
Number of ways - E e, Oo1, Oo2, oO1, oO2 = 5
Probability - 5/9
Case two - For z to be ODD
Number of ways - Eo1, Eo2, eO1, eO2 = 4
Probbility - 4/9
Difference = 1/9
Hope i am using correct approach..