Answer:
Explanation:
We know that between 1 to 10 there are 5 even and 5 odd numbers.
We could get 4 even cards , 4 odd cards or 2 odd and 2 even cards
Let´s check all this combinations
Case 1: When all 4 numbers are even:
We are going to take 4 of the 5 even numbers in the box so we have
![5C4=5](https://img.qammunity.org/2020/formulas/mathematics/college/yx7vsc5kn4eltx8kmwd7409rfthaom2we2.png)
Case 2: When all 4 numbers are odd:
We are going to take 4 of the 5 odd numbers in the box, so we have
![5C4=5](https://img.qammunity.org/2020/formulas/mathematics/college/yx7vsc5kn4eltx8kmwd7409rfthaom2we2.png)
Case 3: When 2 are even and 2 are odd:
We are giong to take 2 from 5 even and odd cards in the box so we have
![5C2 * 5C2](https://img.qammunity.org/2020/formulas/mathematics/college/pdqt8a7iewscilos1oq2iyms2uegd5heju.png)
Remember that we obtain the probability from
So we have the number of favourable outcomes but we need the Total cases for drawing four cards, so we have that:
We are taking 4 of the 10 cards:
![10C_4=210](https://img.qammunity.org/2020/formulas/mathematics/college/ywyldr515t7xpg94qzq16puk5nxjw77eji.png)
Hence we have that the probability that their sum is even
![(5+5+100)/(210)=(11)/(21)](https://img.qammunity.org/2020/formulas/mathematics/college/xs5b5cyq71y0gtujj2d3a1hblqsqqvebjt.png)