In a bag there are three tickets numbered 1, | vedclass

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(A) The total number of ways to draw $4$ tickets with replacement is $3^4 = 81$.Let $S$ be the sum of the $4$ numbers drawn. We want $S$ to be even.Le

Vedclass · Accepted Answer

$\frac{41}{81}$

Vedclass · Answer

(A) The total number of ways to draw $4$ tickets with replacement is $3^4 = 81$.Let $S$ be the sum of the $4$ numbers drawn. We want $S$ to be even.Let $E$ be the event that the sum is even and $O$ be the event that the sum is odd.Let $p_n$ be the probability that the sum of $n$ tickets is even,and $q_n$ be the probability that the sum is odd.For $n=1$,the tickets are ${1, 2, 3}$. The sum is even if we draw $2$ (prob $1/3$) and odd if we draw $1$ or $3$ (prob $2/3$).So,$p_1 = 1/3$ and $q_1 = 2/3$.For $n$ draws,the sum is even if the sum of $n-1$ draws is even and the $n$-th draw is even $(2)$,or if the sum of $n-1$ draws is odd and the $n$-th draw is odd ($1$ or $3$).$p_n = p_{n-1} 	imes (1/3) + q_{n-1} 	imes (2/3)$.Since $q_{n-1} = 1 - p_{n-1}$,we have $p_n = p_{n-1}/3 + (1 - p_{n-1}) 	imes 2/3 = 2/3 - p_{n-1}/3$.For $n=2$: $p_2 = 2/3 - (1/3)/3 = 5/9$.For $n=3$: $p_3 = 2/3 - (5/9)/3 = 13/27$.For $n=4$: $p_4 = 2/3 - (13/27)/3 = 18/27 - 13/81 = 54/81 - 13/81 = 41/81$.Thus,the required probability is $\frac{41}{81}$.

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:

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