Question

A bag contains three tickets numbered $1,2$ and $3$. A ticket is drawn at random and put back in the bag, and this is done four times. The probability that the sum of the numbers drawn is even is

• A. $\frac{40}{81}$
• B. $\frac{41}{81}$
• C. $\frac{14}{27}$
• D. $\frac{13}{81}$

Answer 1

Answer: (B)

$\frac{41}{81}$

We make three cases, namely four odd numbers, two odd numbers and zero odd number; keeping in mind that the sum of four odd numbers & two odd numbers comes out to be even.

Let $X=$ number of odd numbers chosen.

$P$ (sum even) $=P(X=4)+P(X=2)+P(X=0)$

$={\left(\frac{2}{3}\right)}^4{+}^4{C}_2{\left(\frac{2}{3}\right)}^2{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{3}\right)}^4$

$=\frac{41}{81}$