Question

From a box containing 10 cards numbered 1 to10, four cards are drawn together. The probability that their sum is even is

• A. $\frac{1}{2}$
• B. $\frac{11}{21}$
• C. $\frac{10}{21}$
• D. $\frac{1}{21}$

Answer 1

The correct option is B $\frac{11}{21}$

Lets first find the probability of getting a sum as a odd number.
As it ll be more simple as only two cases ll be possible ...
1) if out of 4 cards only 1 is odd
2) 3 odd 1 even
So the total cases for getting odd is $[5C_1\ast 5C_3+5C_3\ast 5C_1]=100$
The number of ways of selecting 4 cards from 10 is ${}^{10}{C}_4=210$
Probability of getting odd $=10/21$
So probability of getting even ll be $11/21$