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3 years ago

Can someone help me in #66

Mathematics

1 answer:

VladimirAG [237]3 years ago

6 0

$\bf log_{{ a}}(xy)\implies log_{{ a}}(x)+log_{{ a}}(y)
\\ \quad \\
% Logarithm of rationals
log_{{ a}}\left( \frac{x}{y}\right)\implies log_{{ a}}(x)-log_{{ a}}(y)
\\ \quad \\
% Logarithm of exponentials
log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\\\\
-------------------------------\\\\
$

$\bf log_4\left( \cfrac{\sqrt{x^5y^7}}{zw^4} \right)\implies log_4(\sqrt{x^5y^7})-log_4(zw^4)
\\\\\\
log_4\left[(x^5y^7)^{^\frac{1}{2}}\right]-log_4(zw^4)\implies 
\cfrac{1}{2}log_4\left[(x^5y^7)\right]-log_4(zw^4)
\\\\\\
\cfrac{1}{2}\left[ log_4(x^5)+log_4(y^7) \right] -[log_4(z)+log_4(w^4)]
\\\\\\
\cfrac{1}{2}log_4(x^5)+\cfrac{1}{2}log_4(y^7)-[log_4(z)+log_4(w^4)]$

$\bf \cfrac{1}{2}\cdot 5log_4(x)+\cfrac{1}{2}\cdot 7log_4(y)-log_4(z)-4log_4(w)
\\\\\\
\cfrac{5}{2}log_4(x)+\cfrac{7}{2}log_4(y)-log_4(z)-4log_4(w)$

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If the object took 4 steps, then it must have gone two steps $\mathrm{N}$ and two steps E, in some permutation. There are $\frac{4 !}{2 ! 2 !}=6$ ways for these four steps of occuring, and the probability is $\frac{6}{4^{4}}$ .

If the object took 6 steps, then it must have gone two steps N and two steps E, and an additional pair of moves that would cancel out, either N / S or W/E. The sequences N, N, N, E, E, S can be permuted in $\frac{6 !}{3 ! 2 ! 1 !}=60$ ways. However, if the first four steps of the sequence are N, N, E, E in some permutation, it would have already reached the point (2,2) in four moves. There are $\frac{4 !}{2 ! 2 !}$ ways to order those four steps and $2 !$ ways to determine the order of the remaining two steps, for a total of 12 sequences that we have to exclude. This gives $60-12=48$ sequences of steps. There are the same number of sequences for the steps N, N, E, E, E, W, so the probability here is $\frac{2 \times 48}{4^{6}}$ .

The total probability is $\frac{6}{4^{4}}+\frac{96}{4^{6}}=\frac{3}{64}$ , and $m+n=067$ .

What is probability?

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