Question 9 (6 marks):
A bag contains five cards, labelled with the letters I, J, K, M and U.
One card is picked at random from the bag and the letter is recorded. Without replacement, another card is picked at random from the bag and the letter is also recorded.
(a) Complete the sample space in the answer space.
(b) By listing all the possible outcomes of the event, find the probability that
(i) the first card picked is labelled with a vowel.
(ii) the first card picked is labelled with a consonant and the second card picked is labelled with a vowel.
Answer:
{(I, J), (I, K), (I, M), (I, U), (J, I), (J, K), (J, M), (J, U), (K, I), (K, J), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )}
Solution:
(a)
S = {(I, J), (I, K), (I, M), (I, U), (J, I), (J, K), (J, M), (J, U), (K, I), (K, J), (K, M), (K, U), (M, I), (M, J), (M, K), (M, U), (U, I), (U, J), (U, K), (U, M)}
(b)(i)
{(I, J), (I, K), (I, M), (I, U), (U, I), (U, J), (U, K), (U, M)}
Probability of first card with vowels=820=25
(b)(ii)
{(J, I), (J, U), (K, I), (K, U), (M, I), (M, U)}
Probability=620=310
A bag contains five cards, labelled with the letters I, J, K, M and U.
One card is picked at random from the bag and the letter is recorded. Without replacement, another card is picked at random from the bag and the letter is also recorded.
(a) Complete the sample space in the answer space.
(b) By listing all the possible outcomes of the event, find the probability that
(i) the first card picked is labelled with a vowel.
(ii) the first card picked is labelled with a consonant and the second card picked is labelled with a vowel.
Answer:
{(I, J), (I, K), (I, M), (I, U), (J, I), (J, K), (J, M), (J, U), (K, I), (K, J), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , )}
Solution:
(a)
S = {(I, J), (I, K), (I, M), (I, U), (J, I), (J, K), (J, M), (J, U), (K, I), (K, J), (K, M), (K, U), (M, I), (M, J), (M, K), (M, U), (U, I), (U, J), (U, K), (U, M)}
(b)(i)
{(I, J), (I, K), (I, M), (I, U), (U, I), (U, J), (U, K), (U, M)}
Probability of first card with vowels=820=25
(b)(ii)
{(J, I), (J, U), (K, I), (K, U), (M, I), (M, U)}
Probability=620=310
Question 10 (6 marks):
(a) Diagram 5.1 shows a Ferris wheel. The distance between point L and point M is 18 m.
Diagram 5.1
Calculate the minimum number of complete spins required to cover the distance of 600 m in a circular motion.
(b) Diagram 5.2 shows one large pizza and two small pizzas. Assume all pizzas are circular with a flat surface.
Diagram 5.2
Using π=227 , calculate the portion of the large pizza which equals to two small pizzas.
Solution:
(a)
Diameter=18 mRadius=9 mCircumference=2πr =2×227×9 m =3967 m3967×rounds(P)=600 mP=600×7396P=35033 roundsP=102033Thus, the minimum number ofcomplete spins required=10.
(b)
Radius of large pizza=14 cmArea of large pizza =πr2=227×(14 cm)2=227×196 cm2=616 cm2Radius of small pizza=7 cmArea of small pizza=πr2=227×(7 cm)2=227×49 cm2=154 cm2Area of 2 small pizza=2×154 cm2=308 cm2Ratio of area of 1 large pizza :2 small pizza616 : 3086162 : 308308 : 308Thus, 12large pizza=2 small pizza
(a) Diagram 5.1 shows a Ferris wheel. The distance between point L and point M is 18 m.
Diagram 5.1
Calculate the minimum number of complete spins required to cover the distance of 600 m in a circular motion.
(b) Diagram 5.2 shows one large pizza and two small pizzas. Assume all pizzas are circular with a flat surface.
Diagram 5.2
Using π=227 , calculate the portion of the large pizza which equals to two small pizzas.
Solution:
(a)
Diameter=18 mRadius=9 mCircumference=2πr =2×227×9 m =3967 m3967×rounds(P)=600 mP=600×7396P=35033 roundsP=102033Thus, the minimum number ofcomplete spins required=10.
(b)
Radius of large pizza=14 cmArea of large pizza =πr2=227×(14 cm)2=227×196 cm2=616 cm2Radius of small pizza=7 cmArea of small pizza=πr2=227×(7 cm)2=227×49 cm2=154 cm2Area of 2 small pizza=2×154 cm2=308 cm2Ratio of area of 1 large pizza :2 small pizza616 : 3086162 : 308308 : 308Thus, 12large pizza=2 small pizza