Probability and Combinatorics: Arrangements, Selections, Solutions

This assignment provides solutions to problems involving permutations, combinations, and probability. It calculates the number of arrangements of letters from a word, determines the number of ways to form teams with specific criteria, and computes the probability of drawing balls of certain colors from a bag without replacement. Formulas for permutations and combinations are applied, and probabilities are calculated based on the given conditions, providing a comprehensive approach to solving these statistical problems. Desklib is the perfect platform for students looking for similar solved assignments and past papers.

QUESTION (a)
If five of the letters from the word INSPIRATION are to be arranged without repetition, then we
calculate the number of different arrangements by applying the formula for permutations without
repetition. That is n!. (Bóna, 2016).
Since, n = 5 and the formula for calculating permutations without repetition is n!
Then:
5! = 120, thus 120 arrangements.
QUESTION b (i)
We know that C(n,r) = n!/r!(n-r)! (Simion & Schmidt, 1985)
Number of ways to select 5 form four students from 8 Form Four students, C(8,5)=56. Team
should consist of 8 students therefore 3 other students have to be Form Five students.
Number of ways to select 3 form five students from 6 students, C(6,3) = 20
Therefore, the total number of teams to be formed so that there are exactly 5 Form Four students
will be; 56 * 20 = 1120
QUESTION b (ii)
For Form Four students to be more than the number of Form Five students, the team must always
have 5, 6 or 7 Form Four students.
Therefore;
C(8,5) = 56
C(8,6) = 28
C(8,7) = 8

For every 5, 6, 7 Form Four students, then there are 3, 2, 1 Form Five students.
Therefore;
C(6,3) = 20
C(6,2) = 15
C(6,1) = 6
Hence:
C(8,5) * C(6,3) = 56 * 20 = 1120
C(8,6) * C(6,2) =28 * 15 = 420
C(8,7) * C(6,1) = 8 * 6 = 48
Total number of teams that can be formed is:
1120 + 420 + 48 = 1588
QUESTION (c)
The bag has 5 red balls (R) and 4 green (G) balls.
Total number of balls in the bag is 5 + 4 = 9
When three balls are drawn from the bag at random without replacement:
Probability that the first, second and third balls drawn without replacement will be red is; R1 =
5/9, R2 = 4/8, R3 = 3/7 respectively while probability that they will be black is; B1 = 4/9, B2 = 3/8,
B3 = 2/7.
This means that (R1, R2, R3 ) = 5/9 * 4/8 * 3/7 = 0.11904 which is probability that all the three
balls will be red
and (B1, B2, B3) = 4/9 * 3/8 * 2/7 = 0.04762 is the probability that all the balls will be black
Therefore, probability that the balls are Red or Black (R or B) will be:

0.11904 + 0.04762 = 0.16666
REFERENCES
Bóna, M. (2016). Combinatorics of permutations. Chapman and Hall/CRC.
Simion, R., & Schmidt, F. W. (1985). Restricted permutations. European Journal of
Combinatorics, 6(4), 383-406.