Question

the chess clubs of two schools consist of, respectively, 8 and 9 players. four members from each club are randomly chosen to participate in a contest between the two schools. the chosen players from one team are then randomly paired with those from the other team, and each pairing plays a game of chess. suppose that rebecca and her sister elise are on the chess clubs at different schools. what is the probability that (a) rebecca and elise will be paired? (b) rebecca and elise will be chosen to represent their schools but will not play each other? (c) either rebecca or elise will be chosen to represent her school?

Answer 1

Using concepts of statistics and Probability, we got 0.055 is the probability  when they play in pair, 0.166 is the probability when they will not play each other and the 0.722 is the probability when either of them will be chosen.

Number of choices to choose players from first school =8!/4!

which is equivalent to =8×7×6×5.

Similarly, Number of choices of how to choose the players from the second school =9!/5! which is equivalent to =9×8×7×6.

a)For each of the 4 boards there are 7×6×5 of choosing teams that have one sister at that table.

Similarly, For each choice of that team there are 8×7×6 choices for the team from the second school that have the other sister at the same board.  

Thus, the probability is  = [(4×7×6×5×8×7×6)] / [(8×7×6×5×9×8×7×6)]

                probability is =(282240)/(5080320)

                 probability is=0.055

b) For each of the 12 different choices of boards for them.  there are 7×6×5 of choosing teams from the  first school,

and 8×7×6 choices for the team from the second school that have the other sisters at the pro scribed boards.

The probability is = [(12×7×6×5×8×7×6)/(8×7×6×5×9×8×7×6)]

      probability is =(846720/5080320)

      probability is =0.006

c)The probability that Elise is chosen is

We know that to choose r items from n items, total number of ways is given by $^nc_r$=$\frac{n!}{r!.(n-r)!}$

For each of the 4 boards there are $^7C_3$ ways of choosing teams that have one sister at that table.

Similarly, to choose the players from the second school =$^9C_4$ ways

We also have 4! internal ways to choose the team members which we have already chosen

The probability the Elise is chosen is

$\frac{^(7C_3).(^9C_4).(4!),(4!)}{8.7.6.5.9.8.7.6}$

=[(7×5×3×7×6×4×3×2×4×3×2)/(8×7×6×5×9×8×7×6)]

=9/18

The probability that Rebecca is chosen is

$\frac{(8c4).(8c3),(4!).(4!)}{8.7.6.5.9.8.7.6}$

=[(7×2×8×7×4×3×2×4×3×2)/(8×7×6×5×9×8×7×6)]

=4/9

The probability that both are chosen is

$\frac{(7c3).(8c3).(4!).(4!)}{8.7.6.5.9.8.7.6}$

[(7×5×8×7×4×3×2×4×3×2)/(8×7×6×5×9×8×7×6)]

=4/18

By inclusion-exclusion the probability either Elise or Rebecca is chosen is

=[(9/18)+(4/9)-(4/18)]

=13/18

=0.722

Hence, probability of Rebecca and Ellise after pairing= 0.055,

probability of Rebecca and Ellise will not paly together=0.006

and, probability that either Rebecca and Ellise play=0.722

To know more about statistics and probability, visit here:

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