Probability

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talaangoshtari: Master | Next Rank: 500 Posts; Posts: 154; Joined: Wed May 21, 2014 4:29 amFollowed by: 1 member; Thanked: 8 times

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

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Source: Beat The GMAT — Problem Solving | Sat Oct 17, 2015 1:34 am

MartyMurray: Legendary Member; Posts: 2135; Joined: Mon Feb 03, 2014 9:26 amFollowed by: 140 members; Location: https://martymurraycoaching.com/; Thanked: 955 times; 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.

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talaangoshtari: Master | Next Rank: 500 Posts; Posts: 154; Joined: Wed May 21, 2014 4:29 amFollowed by: 1 member; Thanked: 8 times

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

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MartyMurray: Legendary Member; Posts: 2135; Joined: Mon Feb 03, 2014 9:26 amFollowed by: 140 members; Location: https://martymurraycoaching.com/; Thanked: 955 times; 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.

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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.

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; 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.

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; 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.

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