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Jan and 5 other children are in a classroom. The principal o

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Jan and 5 other children are in a classroom. The principal o [#permalink]  30 Jul 2018, 11:02
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Question Stats:

62% (00:52) correct 37% (01:19) wrong based on 24 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]  31 Jul 2018, 07:10
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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GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 106

Kudos [?]: 1820 [1] , given: 397

Re: Jan and 5 other children are in a classroom. The principal o [#permalink]  21 Aug 2018, 18:29
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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]  17 Sep 2018, 06:12
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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
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Re: Jan and 5 other children are in a classroom. The principal o [#permalink]  02 Jan 2019, 22:04
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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