BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Probability

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 154
Joined: Wed May 21, 2014 4:29 am
Thanked: 8 times
Followed by:1 members

Probability

by talaangoshtari » Sat Oct 17, 2015 1:34 am
A family with 3 children has one twins. What is the probability that the oldest child is a boy and he is not a twin himself?

A. 1/2
B. 1/3
C. 1/4
D. 3/4
E. 1/5
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 2135
Joined: Mon Feb 03, 2014 9:26 am
Location: https://martymurraycoaching.com/
Thanked: 955 times
Followed by:140 members
GMAT Score:800

by MartyMurray » Sat Oct 17, 2015 8:20 pm
The family has three children, two of which are twins and one of which is not. So seemingly the probability that any of the children is not a twin is 1/3.

The probability of any child being a boy is 1/2.

The probability of any of the three children both not being a twin and being a boy is 1/2 x 1/3 = 1/6.

None of the answer choices match. Hmm. Either I have it wrong or the OA is wrong.
Last edited by MartyMurray on Sat Oct 17, 2015 9:46 pm, edited 1 time in total.
Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 154
Joined: Wed May 21, 2014 4:29 am
Thanked: 8 times
Followed by:1 members

by talaangoshtari » Sat Oct 17, 2015 9:10 pm
Hi Marty Murray,

I had difficulty in understanding the solution of this problem:

S = {([twins], boy), ([twins], girl), (boy, [twins]), (girl, [twins])}

P((boy, [twins])) = 1/4
Top
User avatar
Legendary Member
Posts: 2135
Joined: Mon Feb 03, 2014 9:26 am
Location: https://martymurraycoaching.com/
Thanked: 955 times
Followed by:140 members
GMAT Score:800

by MartyMurray » Sat Oct 17, 2015 9:35 pm
Hmm.

So obviously the probability of a child being a boy is approximately 1/2.

The thing that's a little freakier is the idea that the probability that the older child is not a twin is also 1/2.

Is that idea really right?
Last edited by MartyMurray on Sat Oct 17, 2015 9:54 pm, edited 1 time in total.
Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.
Top
User avatar
Legendary Member
Posts: 2135
Joined: Mon Feb 03, 2014 9:26 am
Location: https://martymurraycoaching.com/
Thanked: 955 times
Followed by:140 members
GMAT Score:800

by MartyMurray » Sat Oct 17, 2015 9:44 pm
Maybe this is the way to look at it.

The mother gives birth twice. She can either have the twins first and the single child second or have the single child first and the twins second.

Maybe we can reword the question this way.

What is the probability that the mother had the twins second and that the other child is a boy?

That's really the same scenario isn't it.

The probability that she had the twins second is 1/2 and the probability that the other child is a boy is 1/2.

So there's our answer, 1/2 x 1/2 = 1/4.
Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.
Top
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

by Matt@VeritasPrep » Mon Oct 26, 2015 12:04 am
I'm thinking of it this way:

P(Twins came second) = 1/2
P(The child that came first is a boy) = 1/2

So our product is 1/2 * 1/2, or 1/4.

We could also use a list of the probabilities:

BBB
BBG
BGB
GBB
BGG
GBG
GGB
GGG

Each of these has TWO possible twin groupings -- BBG has [BB]G and B[BG], for instance -- so we have 16 scenarios of interest. B[BB], B[BG], B[GB], and B[GG] give us what we want, so we have 4 eligible scenarios out of 16, for a probability of 1/4.
Top
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

by Matt@VeritasPrep » Mon Oct 26, 2015 12:10 am
Marty Murray wrote:The family has three children, two of which are twins and one of which is not. So seemingly the probability that any of the children is not a twin is 1/3.
I think the issue here is that we also know that (s)he is the eldest, which tweaks the probability. Since (s)he's the eldest, we either have

Eldest is a twin, i.e. [_ _] _
Eldest is not a twin, i.e. _ [_ _]

These scenarios are equally likely -- at least by the assumptions this problem is certainly making -- so the probability that (s)he is not a twin, given that (s)he is the eldest, is 1/2. Conditional probability gets so funky!

If you're keen, there are scads of fascinating questions on this topic in the second chapter of Blitzstein's Probability.
Top
Post new topic Post Reply
• Page 1 of 1