Question

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

• A. $\frac { 41 } { 81 }$
• B. $\frac { 39 } { 81 }$
• C. $\frac { 40 } { 81 }$
• D. None of these

Answer 1

Answer: (A)

$\frac { 41 } { 81 }$

Answer 2

The total number of ways of selecting 4 tickets $= 3 ^ { 4 } = 81$

The favourable number of ways

= sum of coefficients of $x ^ { 2 } , x ^ { 4 } , \ldots \ldots$ in $\left( x + x ^ { 2 } + x ^ { 3 } \right) ^ { 4 }$

= sum of coefficients of $x ^ { 2 } , x ^ { 4 } , \ldots \ldots$ in $x ^ { 4 } \left( 1 + x + x ^ { 2 } \right) ^ { 4 }$

Let

Then , (On putting $x = 1 )$

and , (On putting $x = - 1 )$

$\therefore 3 ^ { 4 } + 1 = 2 \left( 1 + a _ { 2 } + a _ { 4 } + a _ { 6 } + a _ { 8 } \right)$

$\Rightarrow a _ { 2 } + a _ { 4 } + a _ { 6 } + a _ { 8 } = 41$

Thus sum of the coefficients of

Hence the required probaility