Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
junior college 1 | H1 Maths
One Answer Below
Anyone can contribute an answer, even non-tutors.
Hello I need assistance with (b) & (c). Any help would be appreciated, thank you!
Date Posted: 4 years ago
Views: 552
c)
M stands for Movie ticket.
C stands for Concert ticket.
For the below probabilities, the order of draw is from left to right.
P(at least 2 M are drawn)
= P(2M 2C) + P(3M 1C) + P(4M)
= P(M M C C)
+ P(M C M C)
+ P(M C C M)
+ P(C M M C)
+ P(C M C M)
+ P(C C M M)
+ P(M M M C)
+ P(C M M M)
+ P(M C M M)
+ P(M M C M)
+ P(M M M M)
= 4/10 × 3/9 × 6/8 x 5/7
+ 4/10 × 6/9 × 3/8 x 5/7
+ 4/10 × 6/9 × 5/8 x 3/7
+ 6/10 × 4/9 × 3/8 x 5/7
+ 6/10 × 4/9 × 5/8 x 3/7
+ 6/10 × 5/9 × 4/8 x 3/7
+ 4/10 × 3/9 × 2/8 x 6/7
+ 6/10 × 4/9 × 3/8 x 2/7
+ 4/10 × 6/9 × 3/8 x 2/7
+ 4/10 × 3/9 × 6/8 x 2/7
+ 4/10 × 3/9 × 2/8 x 1/7
= 6(6×5×4×3)/(7×8×9×10) + 4(6×4×3×2)/(7×8×9×10) + (4×3×2×1)/(7×8×9×10)
= (6 × 360 + 4 × 144 + 24)/5040
= 2760/5040
= 23/42
M stands for Movie ticket.
C stands for Concert ticket.
For the below probabilities, the order of draw is from left to right.
P(at least 2 M are drawn)
= P(2M 2C) + P(3M 1C) + P(4M)
= P(M M C C)
+ P(M C M C)
+ P(M C C M)
+ P(C M M C)
+ P(C M C M)
+ P(C C M M)
+ P(M M M C)
+ P(C M M M)
+ P(M C M M)
+ P(M M C M)
+ P(M M M M)
= 4/10 × 3/9 × 6/8 x 5/7
+ 4/10 × 6/9 × 3/8 x 5/7
+ 4/10 × 6/9 × 5/8 x 3/7
+ 6/10 × 4/9 × 3/8 x 5/7
+ 6/10 × 4/9 × 5/8 x 3/7
+ 6/10 × 5/9 × 4/8 x 3/7
+ 4/10 × 3/9 × 2/8 x 6/7
+ 6/10 × 4/9 × 3/8 x 2/7
+ 4/10 × 6/9 × 3/8 x 2/7
+ 4/10 × 3/9 × 6/8 x 2/7
+ 4/10 × 3/9 × 2/8 x 1/7
= 6(6×5×4×3)/(7×8×9×10) + 4(6×4×3×2)/(7×8×9×10) + (4×3×2×1)/(7×8×9×10)
= (6 × 360 + 4 × 144 + 24)/5040
= 2760/5040
= 23/42
The above working can be shortened to :
4!/(2!2!) x 6/10 x 5/9 x 4/8 x 3/7
+ 4!/3! x 4/10 x 3/9 x 2/8 x 6/7
+ 4/10 x 3/9 x 2/8 x 1/7
= [6 x 6 x 5 x 4 x 3 + 4 x 6 x 4 x 3 x 2 + (4 x 3 x 2 x 1] / (7 x 8 x 9 x 10)
= 2760/5040
= 23/42
4!/(2!2!) x 6/10 x 5/9 x 4/8 x 3/7
+ 4!/3! x 4/10 x 3/9 x 2/8 x 6/7
+ 4/10 x 3/9 x 2/8 x 1/7
= [6 x 6 x 5 x 4 x 3 + 4 x 6 x 4 x 3 x 2 + (4 x 3 x 2 x 1] / (7 x 8 x 9 x 10)
= 2760/5040
= 23/42
Alternatively,
c)
P(at least 2 M are drawn)
= 1 - P(1M is drawn) - P(no M is drawn)
= 1 - [ P(M C C C) + P(C C C M) + P(C M C C) + P(C C M C) + P(C C C C)]
= 1 - [4/10 × 6/9 × 5/8 x 4/7 + 6/10 × 5/9 × 4/8 x 4/7 + 6/10 × 4/9 × 5/8 x 4/7 + 6/10 × 5/9 × 4/8 x 4/7 + 6/10 × 5/9 × 4/8 x 3/7]
= 1 - 4(6×5×4×4)/(7×8×9×10) - (6×5×4×3)/(7×8×9×10)
= 1 - (4×480 + 360)/5040
= 1 - 3360/5040
= 1 - 19/42
= 23/42
c)
P(at least 2 M are drawn)
= 1 - P(1M is drawn) - P(no M is drawn)
= 1 - [ P(M C C C) + P(C C C M) + P(C M C C) + P(C C M C) + P(C C C C)]
= 1 - [4/10 × 6/9 × 5/8 x 4/7 + 6/10 × 5/9 × 4/8 x 4/7 + 6/10 × 4/9 × 5/8 x 4/7 + 6/10 × 5/9 × 4/8 x 4/7 + 6/10 × 5/9 × 4/8 x 3/7]
= 1 - 4(6×5×4×4)/(7×8×9×10) - (6×5×4×3)/(7×8×9×10)
= 1 - (4×480 + 360)/5040
= 1 - 3360/5040
= 1 - 19/42
= 23/42
b)
P(4 tickets are not of the same type)
= 1 - P(all 4 tickets are the same type)
= 1 - P(all M) - P(all C)
= 1 - (4/10 × 3/9 × 2/8 × 1/7) (6/10 × 5/9 × 4/8 × 3/7)
= 1 - (4×3×2×1 + 6×5×4×3)/(7×8×9×10)
= 1 - 384/5040
= 1 - 8/105
= 97/105
P(4 tickets are not of the same type)
= 1 - P(all 4 tickets are the same type)
= 1 - P(all M) - P(all C)
= 1 - (4/10 × 3/9 × 2/8 × 1/7) (6/10 × 5/9 × 4/8 × 3/7)
= 1 - (4×3×2×1 + 6×5×4×3)/(7×8×9×10)
= 1 - 384/5040
= 1 - 8/105
= 97/105
See 1 Answer
360 Tutoring Program