Find the probability that a randomly chosen arrangement of the 9 letters in the word CELESTIAL does not have the two Es together. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_s20_qp_53 Year：2020 Question No：7(c)
Answer:
2 Es together \(=\frac{8 !}{2 !}(=20160)\)
Es not together \(=90720-20160=70560\)
Probability \(=\frac{70560}{90720}\)
\(\frac{7}{9}\) or \(0.778\)
Alternative method for question \(7(\) c)
\( { }{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ } \)
\(\frac{7 !}{2 !} \times \frac{8 \times 7}{2}=70560\)
\(7 ! \times k\) in numerator, \(k\) integer \(\geqslant 1\), denominator \(\geqslant 1\)
Multiplying by \({ }^{8} \mathrm{C}_{2} \mathrm{OE}\)
Probability \(=\frac{70560}{90720}\)
\(\frac{7}{9}\) or \(0.778\)
Es not together \(=90720-20160=70560\)
Probability \(=\frac{70560}{90720}\)
\(\frac{7}{9}\) or \(0.778\)
Alternative method for question \(7(\) c)
\( { }{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ }^{\wedge}{ }_{-}{ } \)
\(\frac{7 !}{2 !} \times \frac{8 \times 7}{2}=70560\)
\(7 ! \times k\) in numerator, \(k\) integer \(\geqslant 1\), denominator \(\geqslant 1\)
Multiplying by \({ }^{8} \mathrm{C}_{2} \mathrm{OE}\)
Probability \(=\frac{70560}{90720}\)
\(\frac{7}{9}\) or \(0.778\)
Knowledge points:
5.2.2.1 repetition (e.g. the number of ways of arranging the letters of the word ‘NEEDLESS’)
5.2.2.2 restriction (e.g. the number of ways several people can stand in a line if two particular people must, or must not, stand next to each other). (Questions may include cases such as people sitting in two (or more) rows.) (Questions about objects arranged in a circle will not be included.)
5.3.1 evaluate probabilities in simple cases by means of enumeration of equiprobable elementary events, or by calculation using permutations or combinations (e.g. the total score when two fair dice are thrown.) (e.g. drawing balls at random from a bag containing balls of different colours.)
Solution:
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