A box has 10 cards numbered 1 to 10. If 3 cards are drawn(without replacement), Find probability that sum of numbers on cards drawn is divisible by 3?

Question

Well I have started as below: Total ways of picking cards is 10C3 = 120

All cards in the box are in form
3n - 2 : {1, 4, 7, 10}
3n - 1 : {2, 5, 8}
3n : {3, 6, 9}

Case 1 : All no. are from 3n set then no. of ways is 3C3 = 1
Case 2 : One from 3n-2, 3n-1 and 3n then no. of ways (3n-2)C1 * (3n-1)C1 * (3n)C1
i.e. 4C13C13C1 = 36 or Case 3 : all from 3n-1 set then no. of ways 3C3 = 1

Therefore, total favourable cases = 38
Hence % probability is (38/120)*100 = 31.67%

Is this answer correct? Thank you

Answer 1

You overlooked some cases. The sum of three numbers is divisible by $3$ if

• all three numbers are divisible by $3$, which can occur in $\binom{3}{3} = 1$ way, as you found;
• all three numbers have remainder $1$ when divided by $3$, which can occur in $\binom{4}{3} = 4$ ways;
• all three numbers have remainder $2$ when divided by $3$, which can occur in $\binom{3}{3} = 1$ way;
• one number is divisible by $3$, one number has remainder $1$ when divided by $3$, and one number has remainder $2$ when divided by $3$, which can occur in $\binom{3}{1}\binom{4}{1}\binom{3}{1} = 36$ ways, as you found.

Hence, the number of favorable cases is $1 + 4 + 1 + 36 = 42$.

Since there are $\binom{10}{3} = 120$ possible ways to select three numbers from the set $[10] = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, the probability that the sum of the numbers on the cards that are drawn is divisible by $3$ is $$\frac{42}{120} = \frac{7}{20}$$