JEE Advance - Mathematics (2014 - Paper 2 Offline - No. 5)

The probability that $${x_1} + {x_2} + {x_3}$$ is odd, is

$${{29} \over {105}}$$

$${{53} \over {105}}$$

$${{57} \over {105}}$$

$${{1} \over {2}}$$

Given three Boxes:

JEE Advanced 2014 Paper 2 Offline Mathematics - Probability Question 37 English Explanation

Given, that a card is drow from each Box and $$x_i$$ denote the number on the card drawn from $$i^{\text {tht }}$$ box, $$i=1,2,3$$.

Sample space $$={ }^3 \mathrm{C}_1 .{ }^5 \mathrm{C}_1 .{ }^7 \mathrm{C}_1=3 \cdot 5 \cdot 7=105$$.

Let A be the event when $$x_1+x_2+x_3$$ is odd.

Case I: When all $$x_1, x_2, x_3$$ are odd

No. of ways $$={ }^2 \mathrm{C}_1 .{ }^3 \mathrm{C}_1 .{ }^4 \mathrm{C}_1=2.3 \cdot 4=24$$

Case II: When $$x_1$$ is odd and $$x_2, x_3$$ are even

No. of ways $$={ }^2 \mathrm{C}_1 .{ }^2 \mathrm{C}_1 .{ }^3 \mathrm{C}_1=12$$

Case III: When $$x_2$$ is odd and $$x_1, x_3$$ are even

No. of ways $$={ }^1 \mathrm{C}_1 .{ }^3 \mathrm{C}_1 .{ }^3 \mathrm{C}_1=9$$

Case IV: When $$x_3$$ is odd and $$x_1, x_2$$ are even

No. of ways $$={ }^1 \mathrm{C}_1 .{ }^2 \mathrm{C}_1 . \mathrm{C}_1 \mathrm{C}_1=8$$

No. of favorable cases for A is

$$\begin{aligned} 24+12+9+8 & =53 \\ \text { Required Probability } & =\mathrm{P}(\mathrm{A})=\frac{53}{105} \end{aligned}$$

Hint:

Recall that sum of three numbers is odd only when

(i) All three numbers are odd.

(ii) Two numbers are even and one is odd.