Probability that Sum of Three Numbers Drawn From Boxes is Odd

To find the probability that the sum $x_1 + x_2 + x_3$ is odd, we first analyze the distribution of odd and even numbers in each box:

1. Box 1: Contains $\{1, 2, 3\}$

   • Number of odd cards ($O_1$) = $2$ ($\{1, 3\}$)
   • Number of even cards ($E_1$) = $1$ ($\{2\}$)
   • Total cards = $3$
   • Probabilities: $P(O_1) = \frac{2}{3}$, $P(E_1) = \frac{1}{3}$

2. Box 2: Contains $\{1, 2, 3, 4, 5\}$

   • Number of odd cards ($O_2$) = $3$ ($\{1, 3, 5\}$)
   • Number of even cards ($E_2$) = $2$ ($\{2, 4\}$)
   • Total cards = $5$
   • Probabilities: $P(O_2) = \frac{3}{5}$, $P(E_2) = \frac{2}{5}$

3. Box 3: Contains $\{1, 2, 3, 4, 5, 6, 7\}$

   • Number of odd cards ($O_3$) = $4$ ($\{1, 3, 5, 7\}$)
   • Number of even cards ($E_3$) = $3$ ($\{2, 4, 6\}$)
   • Total cards = $7$
   • Probabilities: $P(O_3) = \frac{4}{7}$, $P(E_3) = \frac{3}{7}$

The total number of possible outcomes when one card is drawn from each box is: $N = 3 \times 5 \times 7 = 105$

The sum $x_1 + x_2 + x_3$ is an odd number if and only if either:

1. All three numbers are odd ($\text{Odd} + \text{Odd} + \text{Odd}$)
2. Exactly one number is odd and two are even ($\text{Odd} + \text{Even} + \text{Even}$)

Let's calculate the number of favorable outcomes for each case:

• Case 1: All three are odd ($O_1, O_2, O_3$) $n(\text{Case 1}) = n(O_1) \times n(O_2) \times n(O_3) = 2 \times 3 \times 4 = 24$

• Case 2: One odd and two even

  • $O_1, E_2, E_3$: $n(O_1) \times n(E_2) \times n(E_3) = 2 \times 2 \times 3 = 12$
  • $E_1, O_2, E_3$: $n(E_1) \times n(O_2) \times n(E_3) = 1 \times 3 \times 3 = 9$
  • $E_1, E_2, O_3$: $n(E_1) \times n(E_2) \times n(O_3) = 1 \times 2 \times 4 = 8$

Summing the favorable outcomes for Case 2: $n(\text{Case 2}) = 12 + 9 + 8 = 29$

Total favorable outcomes: $n(\text{Favorable}) = n(\text{Case 1}) + n(\text{Case 2}) = 24 + 29 = 53$

Thus, the required probability is: $P(x_1 + x_2 + x_3 \text{ is odd}) = \frac{\text{Total Favorable Outcomes}}{\text{Total Possible Outcomes}} = \frac{53}{105}$

Hence, the correct option is B.