Question

A box contains slips numbered 1, 2, 3, 4. Another box contains slips numbered 1, 2, 3. If one slip is taken from each, what is the probability of getting a sum which is a multiple of three?

• A. $\frac{1}{3}$
• B. $1$
• C. $\frac{2}{3}$
• D. $\frac{1}{8}$

Answer 1

The correct option is A $\frac{1}{3}$

Pair of numbers obtained by taking one slip from each box at a time:

(1, 1) (2, 1) (3, 1) (4, 1)

(1, 2) (2, 2) (3, 2) (4, 2)

(1, 3) (2, 3) (3, 3) (4, 3) ---------------(12 Numbers)

Pairs having a sum, which is a multiple of 3 are (2, 1) (1, 2) (4, 2) (3, 3) -------(4 Numbers)

$\therefore$ Probablity = $\frac{4}{12}$ = $\frac{1}{3}$