Some problems of last year divisionals, I need help for

[*Mahi*]

$10.$Prince charming is outside door $A$ and sleeping beauty is in the grey area. There are $5$ doors and the probabilities of doors $A, B, C, D$ and $E$ being open are $0.8, 0.7, 0.6, 0.5$ and $0.4$. What is the probability of Prince Charming being able to get to sleeping beauty?

Using $P+P' = 1$ gives a easy and proper solution here.
1. Door A must be open = $0.8$.
2. Any of door B, door C or (door D and door E) must be open.
2.1. Both D and E open = $0.2$, so one of D or E closed, $0.8$
2.2. B closed = $0.3$
2.3. C closed = $0.4$
2. so, at least one of B, C, (D and E) open = $1 - (0.8 \times 0.3 \times 0.4) = 0.904$
3. So, final probability, $0.8 \times 0.904 = 0.7232$

[Xenon96]

Here is one....
How many four digit even numbers are there so that the sum of their digits is odd???

[Fahim Shahriar]

1st digit(rightmost) has 5 values.

For other digits the cases are:

Case 1: 2nd & 3rd digit are even and 4th is odd. All can have 5 values.
$5 \times 5 \times 5 \times 5=625$

Case 2: 2nd even,3rd odd and 4th even. 4th digit has 4 values.
$5 \times 5 \times 5 \times 4=500$

Case 3: 2nd odd,3rd even and 4th even.
$5 \times 5 \times 5 \times 4=500 $

Case 4: All 3 are odd.
$5 \times 5 \times 5 \times 5=625 $

TOTAL = $2250$

Or you can do it by simple observation. There are $9000$ 4 digit numbers. $4500$ numbers are even. Half of them has sum of digits even and half odd. :

[nafistiham]

Or you can do it by simple observation. There are $9000$ 4 digit numbers. $4500$ numbers are even. Half of them has sum of digits even and half odd. :

I liked the solution
here is another similar.
As the $4^{th}$ digit is even.
The summation of previous digit must be odd.
There are such $\frac {900}{2}=450$ numbers.
For each of them, there are $5$ choices.
So,
\[450 \times 5=2250\]

[Aditi Acharjee]

Or you can do it by simple observation. There are $9000$ 4 digit numbers. $4500$ numbers are even. Half of them has sum of digits even and half odd. :

(y)
Nice solution vaia! But I didn't get the 1st one!!
Btw...I can't solve this problem!!!
For $\triangle ABC$, $\angle C=90^\circ$. $\angle BAC$ is $30^\circ$ and $AB$ is $1\text {cm}$. $D$ is a point within $ABC$ so that $\angle BDC$ is $90^\circ$ and $\angle ACD=\angle DBA$. $AB$ and $CD$ meet at $E$. Find $AE$.

[sowmitra]

Suppose, $D$ is your required point. It lies on the semi-circle on $BC$ (why?), and, $AC$ is tangent to it. Then, $\angle DCA=\angle DBC$ (Tangent-Chord Theorem). So, $\angle EBD=\angle DBC$, which implies, $BD$ is the Angle-Bisector of $\angle ABC$. From this, it follows, $\angle EBC=\angle ECB=60^\circ$, and, $\angle ECA=\angle EAC=30^\circ$. So, $E$ is the mid-point of $AB$.
Therefore, $AE=\boxed{0.5cm}$.

[Fahim Shahriar]

$cos ABC = cos 60° = \frac {BC}{AB}$. $BC=0.5$

$(90° - \angle ACD) + (60° - \angle DBE) = 90°$
$\angle DBE = 60° = \angle BCE$.
$\triangle EBC$ is equilateral. So $AE=BE=BC=0.5$