Answer:
(a) Explained below.
(b) (i) 0, (ii) 0.182, (iii) 0.3182, (iv) 0.2857
Explanation:
The questions are:
a. Illustrate the events. Write the proper symbol for each event. (16 marks)
b. If one student is selected, find the probability of
i. who play an student athletic. soccer and (1 mark)
ii. who play netball but (1 mark) not student athletic.
iii. who student athletic. only in take part (9 marks)
iv. student who take part in netball given that she is not an athletic, or the student play soccer
Solution:
(a)
Total number of students (N) = 110
Number of girls = n (G) = 43, Number of boys = n (B) = 67
Number of athlete student = n (A)= 40
Number of girls and netball = n (G∩N) = 25
Number of girls, netball and athlete = n (G∩A∩N) = 5
Number of boys and soccer = n (B∩S)= 30
Number of boys and athletic = n (B∩A) =15
Number of student supporters = n (SS) = 20
n (B∩S) = 6 × n (G∩A∩N) = 6 × 5 =30
(b)
(i)
P(S∩A) = 0
Since the boys playing soccer only play soccer.
(ii)
Number of students who play netball = n (N) = 25
Number of student who play both netball and athletic = n (N∩A) = 5
Then the number of student only play netball but not athletic,
n (N∩A')= n (N) - n (N∩A) = 20
Then, P (N∩A') = P (N) - P (N∩A) = 20/110 = 0.182
(iii)
Total number of boys taking part in athlete = 15
Total number of girls taking part in athlete = 40-15 =25
Number of girls in athlete and netball = 5
Total number of girls taking part only in athlete = 25-5 =20
P (student only taking part in athlete) = (20 + 15)/110 = 35/110 = 0.3182
(iv)
P (N | A') = P (N∩A')/P(A')
P (N∩A') = 20/110
The total number of students not playing athlete = 110 - 25 - 15 = 70
P(A') = 70/110
The,
P (N | A') = P (N∩A')/P(A') = 20/70 = 0.2857.