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You need to use the definition of conditional probability and make an independence assumption. See my above post.

You need to make an assumption about the dependence of the lives, otherwise there is not enough information to do the question. Since we aren't told anything about it, you probably should assume the lives are independent. Then letting A denote the event that one of them (i.e. at least one of...

It's not wrong (as I said, the second is always correct).

The second one is always correct. The first is only correct if Pr(A and B) = 0 (basically mutually exclusive events).

It's half the answer to the usual coupon collector problem (which is to find the expected number of draws to get at least one of every coupon). In general, it'd be 1/K times the answer to the usual problem, where K is the no. of colours (coupons) available.

$\noindent That didn't come out as an $n$ actually. That's how the $\pi$ symbol looks like in the default font here. If you change to say Times New Roman font (and make the font a bit bigger, say size 3), it looks better, as shown below:$ π.

$\noindent In general, $\frac{d^{2}x}{dt^{2}} \neq \frac{1}{\frac{d^{2}t}{dx^{2}}}$.$

He divided by the determinant of the matrix. Of course, since the original matrix is diagonal, you can invert it by simply inverting (taking the reciprocal of) each diagonal entry.

Yes, a can be positive or negative as written in the question. The equation for the hyperbola doesn't change if you replace a by -a, so you can just find the positive solutions first and negate them if you want the full solution set.

(C)

Q10 is (B).

As written in the question, a can be negative. However, once you get the positive solutions for a, getting the rest of the solutions is trivial since you just negate the positive ones. Since the question was only 2 marks, in my opinion, they should just give the full marks if you get the...

The paper goes till 5 pm.

You want the value of t such that the area of the region in the unit-square that represents the even that "they meet" is equal to 0.5. (Because as it says in (ii), this area is equal to the probability they meet.) So you should essentially find the equations of the relevant lines (in terms...

Those two solution sets are the same (can you see why?). Also, for finding the general solution for cos(x) (or other trig. functions) equal to some "special" values (basically 0 or ±1), it is usually easier to inspect the general solution from the graph rather using the general solution formula...

There are only two standard basis vectors, but there are many different bases. Remember, a basis is not a vector itself, but rather a set of vectors such that everything in the space can be written as a unique linear combination of them. For example, the set {(1, 1), (1, 0)} is another basis...

Because the origin is a point of inflection for the un-translated cubic (y = x3). (Similarly, the origin is the centre of the un-translated circle (x2 + y2 = a2), so (h, k) is the centre of the translated circle (x-h)2 + (y-k)2 = a2. In other words, all you're doing is translating things; the...

Basically the sum of the terms inside the absolute values on the LHS is equal to the expression in the absolute values on the RHS (that is, (a-b)(c-d) + (a-d)(b-c) = (a-c)(b-d)). So the result follows from the triangle inequality (|z| + |w| ≥ |z + w|).

It's true maybe ~50% of the time.

Did you try asking your teacher? Also, I don't think you told us before that that question appeared on a test and you haven't got your score back, so it wasn't ''obvious''.