Bertie

Multiply the first equation by x₁^(1/2), and the second by x₂^(2/3).

Move the negative terms to the other side in each equation, and then align them so that both x₁ terms are on the same side, (the lhs say, the x₂ terms on the rhs).

The product of the two lhs's will equal the product of the two rhs's, do this and simplify.

That should get you 3x₂p₂ = 2x₁p₁

Now solve in conjunction with the third equation.

Here's another 'overlapping' problem.

You have two circles having radii of 15cm and 20cm and they overlap in such a way that at their common points of intersection, the tangents to the two circles are at right angles, (which would imply that the tangent to one circle would pass through the centre of the other).

Calculate their common area, (and it doesn't require calculus).

The triangles have the same angles so they are similar, ratios of corresponding sides will be equal, so 5/2 = x/3, x=15/2 = 7.5.

Nice one Phill -re the 39 runners, two different methods, identical results, can't be bad !

Melody

Instead of using just the colours red and white, suppose that on one card the colours are light red and dark red, on another they are an ever so slightly red and a dull white and on the third a gleaming white and a creamy white.

You draw a card at random and look at just one side.

In how many different ways can this event happen ?

Answer is six. You could be looking at any of the three different reds or any of the three different whites.

Of those six different outcomes how many would you classify as being red (or reddish) ? Answer is three.

Now look at the other sides of these three cards.

If we see light red, the other side is dark red.

If we see dark red the other side will be light red.

If we see an ever so slightly red, the other side will be a dull white.

That's 2 to 1 on the other side being red, and so a probablitiy of 2/3.

(Sorry, Melody's last answer wasn't there when I started typing, but I'll leave mine anyway.)

And BTW, I doubt that my eyesight is sufficiently good to direct a lance successfully !

And I've never ridden a horse.

I'm with Alan, the probability is 2/3.

There are six possible outcomes. I can draw any one of the three cards and I can then see just one of its sides. Imagine a letter on each side of each card, so, card 1 side A, card 1 side B, card 2 side A etc. If I were to draw card number 1, I could see either A or I could  see B, that's two possibilities, not one. Same for the other two cards.

Of the six possible outcomes, three will be red and three will be white.

Of the three possible reds, two will have red the other side one will have white. (We have two possible ways in which the other side is red, one on which the other side is white.) So, the odds are 2 - 1 on the other side being red, a probability of 2/3.

Alan's first post shows this quite clearly.

I'd assumed that the lengths given were the lengths of the vertical lines. What needs clarifying is whether or not the vertical lines are equally spaced, that is, 15/6 cm apart. Even then the area will depend on the actual shape of the boundary curve. Replacing all bits of boundary curve by straight lines (as suggested by Alan) would enable you to get an approximate value.

I agree with your answer Phill.

Anonymous' method is correct as well, (it's probability method or counting method again), but he/she should learn to question results given by a calculator, 1/2016 is absurd.

Anonymous also gets the third question wrong. The question states 'without replacement'. That means that the second probability should be 3/51, not 3/52.

Sorry Phill, but I think that your answer is wrong.

The number of successful outcomes should be 8*C(10,7), not simply C(10,7).

Consider a sequence consisting of three non-yellows followed by seven yellows, the number of ways in which that could occur is 2*2*2*(7 1's) = 8. That is, red or blue, red or blue, red or blue, yellow, yellow, ... , yellow. The numbers will be the same (but in different orders) for each successful sequence.

That gives you a probability of 8*C(10,7)/3^10.

The other way of looking at the problem is through probabilities. Consider the same sequence as above, three non-yellows followed by seven yellows. The probability of that happening is ((2/3)^3)*((1/3)^7) = 8/(3^10). All successful sequences will have the same probability so the total probability will be (8/(3^10))*C(10,7).