Question

19 a form - four class has streams a, b, c and d. the streams are distributed as follows with respect to percentage of antimicrobial resistance (amr) infection.

stream	a	b	c	d
no. of students	45	56	64	60
% with amr	20%	12.5%	25%	10%

(a) a student is chosen at random from the class. determine the:
(i) probability that the student is in form 4d. (2 marks)
(ii) probability that the student is infected with amr. (2 marks)
(b) two students are chosen at random from the school. determine the:
(i) probability that one of the students is in form 4a while the other is in form 4d. (3 marks)
(ii) probability that one of the students is in form 4a while the other is in form 4d and both are infected with amr. (3 marks)

Explanation:

Step1: Calculate total number of students

$45 + 56+64 + 60=225$

Step2: Calculate probability for (a)(i)

The number of students in form 4D is 60. Probability $P(\text{form 4D})=\frac{60}{225}=\frac{4}{15}$

Step3: Calculate number of students with AMR in each stream

In stream A: $45\times0.2 = 9$; In stream B: $56\times0.125 = 7$; In stream C: $64\times0.25=16$; In stream D: $60\times0.1 = 6$

Step4: Calculate total number of students with AMR

$9 + 7+16 + 6=38$

Step5: Calculate probability for (a)(ii)

$P(\text{AMR - infected})=\frac{38}{225}$

Step6: Calculate probability for (b)(i)

$P(\text{one in 4A and one in 4D})=2\times\frac{45}{225}\times\frac{60}{225}=\frac{2\times45\times60}{225\times225}=\frac{8}{25}$

Step7: Calculate number of infected students in 4A and 4D

In 4A: $45\times0.2 = 9$; In 4D: $60\times0.1 = 6$

Step8: Calculate probability for (b)(ii)

$P(\text{one in 4A and one in 4D and both AMR - infected})=2\times\frac{9}{225}\times\frac{6}{225}=\frac{2\times9\times6}{225\times225}=\frac{4}{5625}$

Answer:

(a)(i) $\frac{4}{15}$
(a)(ii) $\frac{38}{225}$
(b)(i) $\frac{8}{25}$
(b)(ii) $\frac{4}{5625}$