JEE Advanced 2014 · Paper 2 · Mathematics Q55

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MathematicsSingle correct+3 / -1Question 55 of 60

Passage

Paragraph For Questions 55 and 56

Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let $x_i$ be the number on the card drawn from the $i^{th}$ box, $i=1,2,3$ .

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

\dfrac{29}{105}

\dfrac{53}{105}

\dfrac{57}{105}

\dfrac12

Answer and worked solution

Official answer: B

Odd counts. Box 1 has 2 odd, 1 even; box 2 has 3 odd, 2 even; box 3 has 4 odd, 3 even. Total $3\times5\times7=105$ outcomes.

Sum odd means one or three odd numbers:

all three odd: $2\cdot3\cdot4=24$ ;
only $x_1$ odd: $2\cdot2\cdot3=12$ ;
only $x_2$ odd: $1\cdot3\cdot3=9$ ;
only $x_3$ odd: $1\cdot2\cdot4=8$ .

Probability. $\dfrac{24+12+9+8}{105}=\dfrac{53}{105}$ .

(B)

Topics:Classical probability

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JEE Advanced 2014 · Paper 2, Mathematics question 55