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secondary 4 | E Maths
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pls help with ii) !!
Date Posted: 3 years ago
Views: 215
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Since it is a fair die, the chances of achieving each number is 1/6.
iia) To end at P, button needs to move 2 steps to the right so the possible combinations are
1. 5,5
2 5,6
3. 6,6
4. 6,5
Hence, P(button ends at P) = (1/6) × (1/6) × 4 = 1/9
• multiply by 4 because there are 4 combinations
iib) To end at Q, the button has to move one time to the left, however, it is impossible to end at Q as the button has to move twice.
Hence, P(button ends at Q) = 0
iic) To end at Y, the button has to move either left then right or right then left. Possible combinations are
1. 1,5
2. 5,1
3. 1,6
4. 6,1
5. 2,5
6. 5,2
7. 2,6
8. 6,2
9. 3,5
10. 5,3
11. 3,6
12. 6,3
13. 4,5
14. 5,4
15. 4,6
16. 6,4
Hence, P(ends at Y) = (1/6) × (1/6) × 16 = 4/9
iid) To end on P or Y or Q, find the sum of parts (iia), (iib) and (iic)
P(end on P, Y or Q) = (1/9) + 0 + (4/9) = 5/9
iia) To end at P, button needs to move 2 steps to the right so the possible combinations are
1. 5,5
2 5,6
3. 6,6
4. 6,5
Hence, P(button ends at P) = (1/6) × (1/6) × 4 = 1/9
• multiply by 4 because there are 4 combinations
iib) To end at Q, the button has to move one time to the left, however, it is impossible to end at Q as the button has to move twice.
Hence, P(button ends at Q) = 0
iic) To end at Y, the button has to move either left then right or right then left. Possible combinations are
1. 1,5
2. 5,1
3. 1,6
4. 6,1
5. 2,5
6. 5,2
7. 2,6
8. 6,2
9. 3,5
10. 5,3
11. 3,6
12. 6,3
13. 4,5
14. 5,4
15. 4,6
16. 6,4
Hence, P(ends at Y) = (1/6) × (1/6) × 16 = 4/9
iid) To end on P or Y or Q, find the sum of parts (iia), (iib) and (iic)
P(end on P, Y or Q) = (1/9) + 0 + (4/9) = 5/9
tysm for the explaination too,,
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