Find the probability that a randomly chosen arrangement of the 10 letters of the word SHOPKEEPER has an $$$\mathrm{E}$$$ at the beginning and an $$$\mathrm{E}$$$ at the end. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mathematics

IGCSE&ALevel

CAIE

Exam No：9709_w20_qp_51 Year：2020 Question No：7(c)

Answer:

Probability $=\frac{\text { Number of ways Es at beginning and end }}{\text { Total number of ways }}$
Probability $=\frac{\frac{8 !}{2 !}}{\frac{10 !}{2 ! \times 3 !}}=\frac{20160}{302400}$
$\frac{1}{15}, 0.0667$
Alternative method for question 7 (c)
Probability $=\frac{3}{10} \times \frac{2}{9}$
$\frac{1}{15}, 0.0667$
Alternative method for question 7 (c)
Probability $=\frac{1}{10} \times \frac{1}{9} \times 3$ !
$\frac{1}{15}, 0.0667$

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.)

Answer:

Knowledge points:

Solution:

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