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Jan and 5 other children are in a classroom. The principal of the scho

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Bunuel
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Jan and 5 other children are in a classroom. The principal of the scho [#permalink] New post   03 Jan 2019, 04:14
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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) \(\frac{1}{2}\)
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Re: Jan and 5 other children are in a classroom. The principal of the scho [#permalink] New post   03 Jan 2019, 04:42
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(1C1 * 5C1)/6C2 = 1/3

So option B

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Re: Jan and 5 other children are in a classroom. The principal of the scho [#permalink] New post   03 Jan 2019, 04:45
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Solution:
Total number of ways of choosing 2 children from 6 children = 6 C 2 = 6!/(4!x2!) = 15 ways
Since the principal of the school will choose two of the children at random and one of them will be Jan, number of ways of another children along with Jan = 5 C 1 = 5 ways
Therefore, probability that Jan will be chosen = 5/15 = 1/3

Alternate Solution:
Given that each children has equal chance of being chosen, when 2 out of 6 children are randomly chosen, then any of 6 children has a chance of 2/6, i.e. 1/3 to be chosen. Therefore, probability that Jan, or any of the other 5 children be chosen is 1/3.

Hence, the Correct Answer is Option B. 1/3
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Re: Jan and 5 other children are in a classroom. The principal of the scho [#permalink] New post   06 Jan 2019, 04:16
Bunuel wrote:
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) \(\frac{1}{2}\)



P of choosing two students out of 6would be 2/6

1/3

IMO B
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Re: Jan and 5 other children are in a classroom. The principal of the scho [#permalink] New post   05 Apr 2023, 09:03
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The principal choose 2 times so possibilities of selecting Jan among 6 are:
(First pick is not Jan and second pick is Jan) or (first pick is Jan and second pick is not Jan)

\(\frac{5}{6}*\frac{1}{5}+\frac{1}{6}*\frac{5}{5}\)
\(\frac{1}{6}+\frac{1}{6}\)
\(\frac{2}{6}\)
\(\frac{1}{3}\)

*selection without replacement

IMO B
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Re: Jan and 5 other children are in a classroom. The principal of the scho [#permalink] New post   05 Apr 2023, 17:22
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Bunuel wrote:
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) \(\frac{1}{2}\)


No need for nPr or cPr. :)

There are six spots Jan could fill:
Chosen
Chosen
Not Chosen
Not Chosen
Not Chosen
Not Chosen

Each of the six is equally likely. Chosen/Total = 2/6 = 1/3.

Answer choice B.
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Re: Jan and 5 other children are in a classroom. The principal of the scho [#permalink] New post   06 Apr 2023, 03:37
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Bunuel wrote:
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) \(\frac{1}{2}\)

Solution:

  • Number of ways to choosing Jan \(= 1\)
  • Number of ways of choosing any one out of 5 other \(= 5C1=5\)

  • Number of ways of choosing any two out of 6 children \(= 6C2=15\)

  • Required probability \(=\frac{1*5C1}{6C2}=\frac{1*5}{15}=\frac{1}{3}\)

Hence the right answer is Option A
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Re: Jan and 5 other children are in a classroom. The principal of the scho [#permalink]
06 Apr 2023, 03:37
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