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Grade 7 POTW In order to find the probability of drawing a winning pair, I must first find out how many possibilities of outcomes there are in total. 4x5=20. Therefore, there are 20 possibilities. Now, I must find out how many of the possibilities are winning numbers. I have made a tree diagram on all of the possibilities and found out that there are 7 winning possibilities. Here are each of the numbers.
Grade 8 POTW: Info: - 12 total symbols using a club, diamond, heart, spade and triangle. - 3 triangles at 3, 6, 12 - After a club must be a diamond or a heart - After a diamond must be a heart or a spade - After a heart must be a spade or a triangle - After a spade must be a triangle or a club - After a triangle must be a club or a diamond.
Since the example gives us one valid sequence, we know for sure that this isn’t a trick question and there’s at least one possible solution :)))))
We should probably use trial and error for this question. And we need to work backwards for this question. For the first two positions, we need to use the 3rd symbol. There are 4 possible ways, which are club-heart, diamond-heart, diamond-spade, heart-spade.
After the triangle in spot 3, we only have two options for spot 4. These would be the club or a diamond. Since there is also a triangle in position 6, there are only a total of 4 possibilites for spots 4 and 5. But since one of these cannot work (club-spade) because the club must be followed by something other than the spade, there are only 3 options for spots 4 and 5. Now there are 4 x 3 = 12 total possibilties so far.
Since we do not have a triangle from spot 7 all the way to spot 11, we need to look at each possibility. There also cannot be a triangle in any of these spots, so this will get rid of that possibility. We can start off with what can follow a triangle. Only a club or a diamond can follow. Now we can use that and go from there. First, the possibilities that have a diamond starting: - DHSCH, DSCDS, DSCDH, DSCHS. Club: - CDSCH, CHSCH
This is a total of 6 possitiblities. Now we need to multiply the previous 12 by the new 6. 12 x 6 = 72.
Grade 8 POTW We know that position 3, 6 and 12 is all triangles. We can use this to find out all possible other positions. The total number of valid sequences is obtained by multiplying the number of options together. This can be done by solving for the positions beside each triangle. We can separate them into positions 1 and 2, 4 and 5, and 7, 8, 9, 10 and 11. Position 1 and 2: Only a heart and a spade can appear before a triangle. Therefore, there are 2 possible options at position 2. If it is a heart, then position 1 can be a diamond or a club. If it is a spade, then position 1 can be a heart or diamond. Therefore, there are 4 possible options for both position 1 and 2. Position 4 and 5: This is a little different from the first way. We know that a club or diamond must be placed at position 4. A heart or spade must be placed at position 5. If position 4 was a club, then position 5 must be a diamond or a heart. It can’t be a diamond so it must be a spade. If position 4 was a diamond, then a heart or spade can follow giving 3 total options for position 4 and 5. Position 7, 8, 9, 10 and 11: We know that position 7 is a club or a diamond and position 11 is a heart or spade. We will try all the situations. If position 7 was a club, then position 8 must be a diamond or heart. If it were a diamond, then position 9 must be a heart or a spade. If position 9 was a heart, then position 10 has to be a spade or a triangle. If it were a spade, then position 11 must be a triangle or club. Both don’t work. Position 10 can also not be a triangle as the only 3 are known already. If position 9 were a spade, then position 10 must be a triangle or club (or just club). If 10 were a club, 11 must be a diamond or heart. It can be a heart, but not a diamond. If position 8 were a heart, then position 9 must be a spade or triangle (or just spade). If position 9 was a spade, then 10 has to be a triangle or club (or just club). If position 10 was a club, then 11 must be a diamond or heart. It can only be diamond. If position 7 was a diamond, then position 8 must be a heart or a spade. If it were a heart, then position 9 must be a spade or triangle (or just spade). If position 9 were a spade, then position 10 must be a triangle or club (or just club). If position 10 were a club, then 11 must be a diamond or a heart. Only a heart works. If position 8 was a spade, then position 9 would be a triangle or club (or just club). If it were a club, then 10 must be a diamond or a heart. If 10 were a diamond, then 11 would be a heart or a spade which both works. If 10 were a heart, then 11 must be a spade or a triangle. Only a spade works. Both of these together adds up to 6 options. If we multiply everything together, we get a total of 4 x 3 x 6 = 72 total valid sequences. Therefore, there are 72 valid sequences.
Grade 8 POTW We know that position 3, 6 and 12 is all triangles. We can use this to find out all possible other positions. The total number of valid sequences is obtained by multiplying the number of options together. This can be done by solving for the positions beside each triangle. We can separate them into positions 1 and 2, 4 and 5, and 7, 8, 9, 10 and 11. Position 1 and 2: Only a heart and a spade can appear before a triangle. Therefore, there are 2 possible options at position 2. If it is a heart, then position 1 can be a diamond or a club. If it is a spade, then position 1 can be a heart or diamond. Therefore, there are 4 possible options for both position 1 and 2. Position 4 and 5: This is a little different from the first way. We know that a club or diamond must be placed at position 4. A heart or spade must be placed at position 5. If position 4 was a club, then position 5 must be a diamond or a heart. It can’t be a diamond so it must be a spade. If position 4 was a diamond, then a heart or spade can follow giving 3 total options for position 4 and 5. Position 7, 8, 9, 10 and 11: We know that position 7 is a club or a diamond and position 11 is a heart or spade. We will try all the situations. If position 7 was a club, then position 8 must be a diamond or heart. If it were a diamond, then position 9 must be a heart or a spade. If position 9 was a heart, then position 10 has to be a spade or a triangle. If it were a spade, then position 11 must be a triangle or club. Both don’t work. Position 10 can also not be a triangle as the only 3 are known already. If position 9 were a spade, then position 10 must be a triangle or club (or just club). If 10 were a club, 11 must be a diamond or heart. It can be a heart, but not a diamond. If position 8 were a heart, then position 9 must be a spade or triangle (or just spade). If position 9 was a spade, then 10 has to be a triangle or club (or just club). If position 10 was a club, then 11 must be a diamond or heart. It can only be diamond. If position 7 was a diamond, then position 8 must be a heart or a spade. If it were a heart, then position 9 must be a spade or triangle (or just spade). If position 9 were a spade, then position 10 must be a triangle or club (or just club). If position 10 were a club, then 11 must be a diamond or a heart. Only a heart works. If position 8 was a spade, then position 9 would be a triangle or club (or just club). If it were a club, then 10 must be a diamond or a heart. If 10 were a diamond, then 11 would be a heart or a spade which both works. If 10 were a heart, then 11 must be a spade or a triangle. Only a spade works. Both of these together adds up to 6 options. If we multiply everything together, we get a total of 4 x 3 x 6 = 72 total valid sequences. Therefore, there are 72 valid sequences.
Grade 7 POTW Red Cards: 1, 3, 5, 9, 12 Blue Cards: 1, 3, 4, 5 First, we’ll find out the total number of possible pairs. 5 x 4 = 20. Out of all these options, the only perfect square pairs are: (1, 1), (1, 4), (3, 3), (5, 5), (9, 1), (9, 4), and (12, 3). That’s 7/20 = 0.35 = 35% Therefore, the probability of getting a winning pair is 7/20, 35%, or 0.35.
Grade 7 POTW
ReplyDeleteIn order to find the probability of drawing a winning pair, I must first find out how many possibilities of outcomes there are in total. 4x5=20. Therefore, there are 20 possibilities. Now, I must find out how many of the possibilities are winning numbers. I have made a tree diagram on all of the possibilities and found out that there are 7 winning possibilities. Here are each of the numbers.
1x1=1 1
1x4=4 2
3x3=9 3
5x5=25 4
1x9=9 5
4x9=36 6
12x3=36 7
Therefore, the probability of drawing a winning ticket is 7/20, 35% or 0.35.
Grade 8 POTW:
ReplyDeleteInfo:
- 12 total symbols using a club, diamond, heart, spade and triangle.
- 3 triangles at 3, 6, 12
- After a club must be a diamond or a heart
- After a diamond must be a heart or a spade
- After a heart must be a spade or a triangle
- After a spade must be a triangle or a club
- After a triangle must be a club or a diamond.
Since the example gives us one valid sequence, we know for sure that this isn’t a trick question and there’s at least one possible solution :)))))
We should probably use trial and error for this question. And we need to work backwards for this question. For the first two positions, we need to use the 3rd symbol. There are 4 possible ways, which are club-heart, diamond-heart, diamond-spade, heart-spade.
After the triangle in spot 3, we only have two options for spot 4. These would be the club or a diamond. Since there is also a triangle in position 6, there are only a total of 4 possibilites for spots 4 and 5. But since one of these cannot work (club-spade) because the club must be followed by something other than the spade, there are only 3 options for spots 4 and 5. Now there are 4 x 3 = 12 total possibilties so far.
Since we do not have a triangle from spot 7 all the way to spot 11, we need to look at each possibility. There also cannot be a triangle in any of these spots, so this will get rid of that possibility. We can start off with what can follow a triangle. Only a club or a diamond can follow. Now we can use that and go from there. First, the possibilities that have a diamond starting:
- DHSCH, DSCDS, DSCDH, DSCHS.
Club:
- CDSCH, CHSCH
This is a total of 6 possitiblities. Now we need to multiply the previous 12 by the new 6. 12 x 6 = 72.
There are 72 sequences that will work.
Fiona Bian
Grade 8 POTW
ReplyDeleteWe know that position 3, 6 and 12 is all triangles. We can use this to find out all possible other positions. The total number of valid sequences is obtained by multiplying the number of options together. This can be done by solving for the positions beside each triangle. We can separate them into positions 1 and 2, 4 and 5, and 7, 8, 9, 10 and 11.
Position 1 and 2:
Only a heart and a spade can appear before a triangle. Therefore, there are 2 possible options at position 2. If it is a heart, then position 1 can be a diamond or a club. If it is a spade, then position 1 can be a heart or diamond. Therefore, there are 4 possible options for both position 1 and 2.
Position 4 and 5:
This is a little different from the first way. We know that a club or diamond must be placed at position 4. A heart or spade must be placed at position 5. If position 4 was a club, then position 5 must be a diamond or a heart. It can’t be a diamond so it must be a spade. If position 4 was a diamond, then a heart or spade can follow giving 3 total options for position 4 and 5.
Position 7, 8, 9, 10 and 11:
We know that position 7 is a club or a diamond and position 11 is a heart or spade. We will try all the situations.
If position 7 was a club, then position 8 must be a diamond or heart. If it were a diamond, then position 9 must be a heart or a spade. If position 9 was a heart, then position 10 has to be a spade or a triangle. If it were a spade, then position 11 must be a triangle or club. Both don’t work. Position 10 can also not be a triangle as the only 3 are known already. If position 9 were a spade, then position 10 must be a triangle or club (or just club). If 10 were a club, 11 must be a diamond or heart. It can be a heart, but not a diamond. If position 8 were a heart, then position 9 must be a spade or triangle (or just spade). If position 9 was a spade, then 10 has to be a triangle or club (or just club). If position 10 was a club, then 11 must be a diamond or heart. It can only be diamond.
If position 7 was a diamond, then position 8 must be a heart or a spade. If it were a heart, then position 9 must be a spade or triangle (or just spade). If position 9 were a spade, then position 10 must be a triangle or club (or just club). If position 10 were a club, then 11 must be a diamond or a heart. Only a heart works. If position 8 was a spade, then position 9 would be a triangle or club (or just club). If it were a club, then 10 must be a diamond or a heart. If 10 were a diamond, then 11 would be a heart or a spade which both works. If 10 were a heart, then 11 must be a spade or a triangle. Only a spade works.
Both of these together adds up to 6 options.
If we multiply everything together, we get a total of 4 x 3 x 6 = 72 total valid sequences.
Therefore, there are 72 valid sequences.
Grade 8 POTW
ReplyDeleteWe know that position 3, 6 and 12 is all triangles. We can use this to find out all possible other positions. The total number of valid sequences is obtained by multiplying the number of options together. This can be done by solving for the positions beside each triangle. We can separate them into positions 1 and 2, 4 and 5, and 7, 8, 9, 10 and 11.
Position 1 and 2:
Only a heart and a spade can appear before a triangle. Therefore, there are 2 possible options at position 2. If it is a heart, then position 1 can be a diamond or a club. If it is a spade, then position 1 can be a heart or diamond. Therefore, there are 4 possible options for both position 1 and 2.
Position 4 and 5:
This is a little different from the first way. We know that a club or diamond must be placed at position 4. A heart or spade must be placed at position 5. If position 4 was a club, then position 5 must be a diamond or a heart. It can’t be a diamond so it must be a spade. If position 4 was a diamond, then a heart or spade can follow giving 3 total options for position 4 and 5.
Position 7, 8, 9, 10 and 11:
We know that position 7 is a club or a diamond and position 11 is a heart or spade. We will try all the situations.
If position 7 was a club, then position 8 must be a diamond or heart. If it were a diamond, then position 9 must be a heart or a spade. If position 9 was a heart, then position 10 has to be a spade or a triangle. If it were a spade, then position 11 must be a triangle or club. Both don’t work. Position 10 can also not be a triangle as the only 3 are known already. If position 9 were a spade, then position 10 must be a triangle or club (or just club). If 10 were a club, 11 must be a diamond or heart. It can be a heart, but not a diamond. If position 8 were a heart, then position 9 must be a spade or triangle (or just spade). If position 9 was a spade, then 10 has to be a triangle or club (or just club). If position 10 was a club, then 11 must be a diamond or heart. It can only be diamond.
If position 7 was a diamond, then position 8 must be a heart or a spade. If it were a heart, then position 9 must be a spade or triangle (or just spade). If position 9 were a spade, then position 10 must be a triangle or club (or just club). If position 10 were a club, then 11 must be a diamond or a heart. Only a heart works. If position 8 was a spade, then position 9 would be a triangle or club (or just club). If it were a club, then 10 must be a diamond or a heart. If 10 were a diamond, then 11 would be a heart or a spade which both works. If 10 were a heart, then 11 must be a spade or a triangle. Only a spade works.
Both of these together adds up to 6 options.
If we multiply everything together, we get a total of 4 x 3 x 6 = 72 total valid sequences.
Therefore, there are 72 valid sequences.
Grade 7 POTW
ReplyDeleteRed Cards: 1, 3, 5, 9, 12
Blue Cards: 1, 3, 4, 5
First, we’ll find out the total number of possible pairs. 5 x 4 = 20. Out of all these options, the only perfect square pairs are:
(1, 1), (1, 4), (3, 3), (5, 5), (9, 1), (9, 4), and (12, 3).
That’s 7/20 = 0.35 = 35%
Therefore, the probability of getting a winning pair is 7/20, 35%, or 0.35.