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Bunuel
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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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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Bunuel
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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Bunuel
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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P(JO) where J is Jan and O is other children
= 1/6*5/5*2!(arrangement since jan can be choosen first or second )
= 1/3
Answer is B
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