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probability question (1 Viewer)
- Thread starter apollo1
- Start date
for i) we have 5 lines which are not parallel, so they all intersect, and no three lines are concurrent, so no 3 lines intersect in one point.
This means the first line will intersect the other lines 4 times...
The second line will then intersect the remaining 3 lines...
The third line will then intersect the remaining 2 lines....and
The fourth line will then intersect the last line...
This means there are 4+3+2+1= 10 points of intersections.
For ii)
If you draw a diagram you can see that every line has 4 intersection points....so this means if we want to choose 3 random points so that they lie on one of the given lines, we can choose 3 points of intersection from the 4 points of intersection each line has in:
ways.
Now that is for one line, but remember there are 5 lines...so the number of ways of choosing 3 points from 4 points for all the 5 lines is 5 x 4 = 20 ways.
hence:
= \frac{20}{\binom{10}{3}}= \frac{1}{6})
remembering the sample space is choosing 3 points from 10.
I'm pretty sure the second part is right, but you never know with probability, especially a question like this...which is nearly the same as question 8 of 1990/1 i think
This means the first line will intersect the other lines 4 times...
The second line will then intersect the remaining 3 lines...
The third line will then intersect the remaining 2 lines....and
The fourth line will then intersect the last line...
This means there are 4+3+2+1= 10 points of intersections.
For ii)
If you draw a diagram you can see that every line has 4 intersection points....so this means if we want to choose 3 random points so that they lie on one of the given lines, we can choose 3 points of intersection from the 4 points of intersection each line has in:
Now that is for one line, but remember there are 5 lines...so the number of ways of choosing 3 points from 4 points for all the 5 lines is 5 x 4 = 20 ways.
hence:
I'm pretty sure the second part is right, but you never know with probability, especially a question like this...which is nearly the same as question 8 of 1990/1 i think
largarithmic
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- 2011
(i) Each pair of lines intersects once (as no pair are parallel) and each point of intersection has exactly two lines through it (as there are no concurrent triples of lines). Hence the number of points of intersection is exactly 5C2 = 10.
(ii) Note that each line has exactly 4 points of intersection on it; corresponding to its intersection with each of the other four lines. This means that the number of triples of intersection points lying on the same line is 5 * (4C3), with the 5 to choose the line they lie on and the 4C3 to choose the actual triple. The total number of triples is obviously 10C3. The result is thus, probability = 5(4C3)/10C3 = 5*4/120 = 1/6
(ii) Note that each line has exactly 4 points of intersection on it; corresponding to its intersection with each of the other four lines. This means that the number of triples of intersection points lying on the same line is 5 * (4C3), with the 5 to choose the line they lie on and the 4C3 to choose the actual triple. The total number of triples is obviously 10C3. The result is thus, probability = 5(4C3)/10C3 = 5*4/120 = 1/6
michaeljennings
Active Member
I cant believe I have to compete with you guys lol...you guys are heaps good at ext 2
First line has no intersections
Second has one
Third has two....
Total = 1 + 2 + 3 + 4 = 10
P(1 on same line) = 1
P(2 on the same line (there are two potential lines)) = 6/9 (just count
) = 2/3
P(3 on that line) = 2 / 8 = 1/4
P(3) = 1 * 1/4 * 2/3 = 1/6
Second has one
Third has two....
Total = 1 + 2 + 3 + 4 = 10
P(1 on same line) = 1
P(2 on the same line (there are two potential lines)) = 6/9 (just count
) = 2/3P(3 on that line) = 2 / 8 = 1/4
P(3) = 1 * 1/4 * 2/3 = 1/6
michaeljennings
Active Member
oh i see hahaha, what marks did you get for ext 1 and ext 2 in the hsc?
i did 3u in yr 11 and i got 48/50 which i was happy with..but i was aiming for 98+ in 4u and only got 96 lol...stuffed up a few easy marks last yr