I don't know how to explain why logM is reasonable approximation to the one of the roots.
The only thing I could come up with is lnMX=lne^x and so x=lnM + ln X and if M gets too large the ln X is very small comparing to x itself. But, I'm not sure if it's sensible at all. If substitute x=lnM you get M.lnM=M And so M=e?
When M=e, the approximation is perfect. I think one of the ways to do it is to consider the function e^x - Mx. Specifically, are there are any turning points and what is the gradient like around those turning points?
When M=e, the approximation is perfect. I think one of the ways to do it is to consider the function e^x - Mx. Specifically, are there are any turning points and what is the gradient like around those turning points?
I don't want to cheat and use wolfram.
?
EDIT: it should be 1 instead of e, and the bottom curve shouldn't touch the y-axis. And I can't follow why it's a perfect approximation? This approximation means that M can only be e?
I plotted them on the graph and said that, as M gets larger, one of the roots get closer and closer to 0. Then, consider mx=e^x at x=1 if M is large enough Mx>e^x, but as x gets bigger, no matter how big the M is, we'll get mx=e^x and then e^x>mx
The only thing I could come up with is lnMX=lne^x and so x=lnM + ln X and if M gets too large the ln X is very small comparing to x itself.
And if lnM is a reasonable approximation of the root, we get M=e therefore only one possible value of M. But clearly, M can be any big number. What's wrong?
And if lnM is a reasonable approximation of the root, we get M=e therefore only one possible value of M. But clearly, M can be any big number. What's wrong?
How did you get this, nobody said w=log(M)? What has been suggested is that w≈log(M). You are also told that w>1, which isn't the case when M=e. Otherwise, the equation will have one solution where the straight line is tangent to the curve.
How did you get this, nobody said w=log(M)? What has been suggested is that w≈log(M). You are also told that w>1, which isn't the case when M=e. Otherwise, the equation will have one solution where the straight line is tangent to the curve.
If lnMX=lne^x and so x=lnM + ln X. And bigger M means bigger root. So as x get's larger difference between x and ln x get's bigger and you can ignore it, right? If so, when M is huge, it's pretty good approximation. The problem is, I tried M=10^35, 10^75. And lnM doesn't seem to be a good approximation for x. So, there is a problem somewhere?
Do you think, it's not acceptable to explain it in words?
To be honest, I don't know what is acceptable and what is not. Similarly, I don't think they're looking for something too specific, as many people have done only C1-C4 (+ FP1) at the time of the interviews.
There is a pile of 129 coins on a table, all unbiased except for one which has heads on both sides. Bobchooses a coin at random and tosses it eight times. The coin comes up heads every time. What is the probability that it will come up heads the ninth time as well?
P(coin is unbiased) = 128/129 P(coin is biased) = 1/129 P(unbiased coin is heads 8 times) = (1/2)^8 = 1/256 P(biased coin is heads 8 times) = 1
So the probability of an unbiased coin doing what has been done is 1/258
And the probability of a biased coin doing what is done is 2/258
So the probability we have a biased coin is 2/3 and the probability we have a unbiased coin is 1/3
Then finally the probability it will be heads again is (2/3)(1) + (1/3)(1/2) = 5/6
I don't really know if this is correct, and sorry I couldn't use proper notation - I've never really done probability except a tiny bit in S1 so I've just done this out of logic alone.
P(coin is unbiased) = 128/129 P(coin is biased) = 1/129 P(unbiased coin is heads 8 times) = (1/2)^8 = 1/256 P(biased coin is heads 8 times) = 1
So the probability of an unbiased coin doing what has been done is 1/258
And the probability of a biased coin doing what is done is 2/258
So the probability we have a biased coin is 2/3 and the probability we have a unbiased coin is 1/3
Then finally the probability it will be heads again is (2/3)(1) + (1/3)(1/2) = 5/6
I don't really know if this is correct, and sorry I couldn't use proper notation - I've never really done probability except a tiny bit in S1 so I've just done this out of logic alone.
Oops, didn't actually check what you were doing there, just assumed you were essentially restating the previous bit What you wrote it fine I think. (For the record: P(A|B)=P(AnB)/P(B) is a handy formula for this kinda thing, to make it clearer what's going on).
P(coin is unbiased) = 128/129 P(coin is biased) = 1/129 P(unbiased coin is heads 8 times) = (1/2)^8 = 1/256 P(biased coin is heads 8 times) = 1
So the probability of an unbiased coin doing what has been done is 1/258
And the probability of a biased coin doing what is done is 2/258
So the probability we have a biased coin is 2/3 and the probability we have a unbiased coin is 1/3
Then finally the probability it will be heads again is (2/3)(1) + (1/3)(1/2) = 5/6
I don't really know if this is correct, and sorry I couldn't use proper notation - I've never really done probability except a tiny bit in S1 so I've just done this out of logic alone.