Lời giải:

Gọi $X$ là biến cố sinh viên là nữ và $Y$ là biến cố sinh viên học ngành A.

Theo bài ra ta có:

$P(X)=0,52; P(Y) = 0,05; P(XY)=0,02$
a. 

Xác suất sinh viên là nữ nếu sinh viên học ngành A là:
$P(X|Y) = \frac{P(XY)}{P(Y)}=\frac{0,02}{0,05}=\frac{2}{5}$

b. 

Xác suất sinh viên học ngành A biết sinh viên là nữ là:

$P(Y|X) = \frac{P(XY)}{P(X)}=\frac{0,02}{0,52}=\frac{1}{26}$

Lời giải:

Gọi $S=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^n}$

$5S=1+\frac{1}{5}+....+\frac{1}{5^{n-1}}$

$\Rightarrow 4S=5S-S=1-\frac{1}{5^n}$

$S=\frac{1}{4}-\frac{1}{4.5^n}$

Khi $n\to \infty$ thì $4.5^n\to \infty\Rightarrow \frac{1}{4.5^n}\to 0$

$\Rightarrow \lim S = \frac{1}{4}-\lim \frac{1}{4.5^n}=\frac{1}{4}$

1+tan^2 x = 1/cos^2 x

=> 1+ t^2 = 1/cos^2 x

=> 3 + 3t^2 = 3/cos^2 x

PT TRỞ THÀNH :

3 + 3t^2 - 2t + 1 = 0

<=> 3t^2 - 2t + 4 = 0

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