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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # Jan and 5 other children are in a classroom. The principal o  Question banks Downloads My Bookmarks Reviews Important topics
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Retired Moderator Joined: 07 Jun 2014
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Jan and 5 other children are in a classroom. The principal o [#permalink]
Expert's post 00:00

Question Stats: 46% (00:57) correct 53% (00:30) wrong based on 43 sessions
Jan and 5 other children are in a classroom. The principal of the school will choose two of the children at random. What is the probability that Jan will be chosen?

(A) $$\frac{4}{5}$$
(B) $$\frac{1}{3}$$
(C) $$\frac{2}{5}$$
(D) $$\frac{7}{15}$$
(E) $$\frac{1}{2}$$
[Reveal] Spoiler: OA

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Sandy
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Re: Jan and 5 other children are in a classroom. The principal o [#permalink]
There are altogether 6 children hence possible outcome = 6*5 = 30
If Jan is to be chosen.
Then Either Jan is chosen and someone else is chosen or Someone else is chosen first and then Jan is chosen
hence possible outcome = 1*5 + 5*1 = 10

probability = 10/30 =1/3
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This is my response to the question and may be incorrect. Feel free to rectify any mistakes
Need Practice? 20 Free GRE Quant Tests available for free with 20 Kudos Retired Moderator Joined: 07 Jun 2014
Posts: 4808
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 145

Kudos [?]: 2288  , given: 393

Re: Jan and 5 other children are in a classroom. The principal o [#permalink]
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Expert's post
Explanation

The probability of any event equals the number of ways to get the desired outcome divided by the total number of outcomes.

Start with the denominator, which is the total number of ways that the principal can choose two children from the classroom. Use the fundamental counting principle. There are 6 possible options for the first choice and 5 for the second, giving (6)(5) = 30 possibilities.

However, this double-counts some cases; for example, choosing Jan and then Robert is the same as choosing Robert and then Jan.

Divide the total number of pairs by 2: $$\frac{6 \times 5}{2}= 1$$. Alternatively, use the formula for a set in which the order doesn’t matter: $$\frac{total!}{in! \times out!}$$.

In this case: $$\frac{6!}{2! \times 4!}=15$$.

Now compute the numerator, which is the number of pairs that include Jan. Since the pair only includes two children and one is already decided (Jan), there are exactly 5 options for the other child. Thus, there are 5 total pairs that include Jan: Jan with each of the other students.

The probability of choosing a pair with Jan is $$\frac{1}{3}$$.
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Re: Jan and 5 other children are in a classroom. The principal o [#permalink]
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Is this method right?

Probablility of Jan not being chosen: 5/6 *4/5=4/6=2/3
Probability of Jan being chosen = 1 - 2/3=1/3
Manager  Joined: 01 Nov 2018
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Re: Jan and 5 other children are in a classroom. The principal o [#permalink]
Expert's post
kruttikaaggarwal wrote:
Is this method right?

Probablility of Jan not being chosen: 5/6 *4/5=4/6=2/3
Probability of Jan being chosen = 1 - 2/3=1/3

Yes! much simpler too! Thanks !! Re: Jan and 5 other children are in a classroom. The principal o   [#permalink] 02 Jan 2019, 22:04
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