1i misread that as x≤y.
\hspace{-16}$If $\mathbf{1\leq x<y\leq 100}$ and $\mathbf{x,y\in\mathbb{Z}}$. Then find the Probability that $\mathbf{i^x+i^y\in\mathbb{R}}$\\\\ Where $\mathbf{i=\sqrt{-1}}$
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5 Answers
1is it 38 ?
n(s) = 100*100
n(E) = 2cases;
case-1: x and y are both even => 50*50
case-2: when 1 is i and the other is -i => 2*25*25
=> probaility = 3/8.
1057first let us count the total no of cases...
when x is 1 y can have 99 values...
when x is 2 y can have 98 values...
therefore total no of cases is :
\sum_{n=1}^{99}{n} = 99 x 50....
case 1:x and y are both even...
for x=2 y can have 49 values
for x=4 y can have 48 values and so on....
therefore favourable cases is \sum_{n=1}^{49}{n}=49 x 25
case 2:x is of the form 4n+1 and y of 4n+3(den de expression will result in 0)
for x=1 y can have 33 values....
for x=5 y can have 32 values and so on...
favourable cases is \sum_{n=1}^{33}{n} = 33 x 17
total favourable cases is 33 x 17 + 49 x 25 = 1786
total cases = 4950
therefore probability = 17864950
1057@rishabh can u explain how the total no of cases turn out to be 100 x 100.....are you following the condition x<y???
1057even if u misread it the total no of (x,y) pairs will be 5050 and not 10000....
so wat's de correct answer????
i wrote a little bit wrong...
in the 2nd case it will be \sum_{n=1}^{25}{n} = 25 x 13
therefore probability will be 3199
