Probability Ques 11
Passage Based Problems
Box I contains three cards bearing numbers $1,2,3$; box II contains five cards bearing numbers $1,2,3,4,5$; and box III 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 $i=1,2,3$.
(2014 Adv.)
The probability that $x _1+x _2+x _3$ is odd, is
(a) $\frac{29}{105}$
(b) $\frac{53}{105}$
(c) $\frac{57}{105}$
(d) $\frac{1}{2}$
Show Answer
Answer:
Correct Answer: 11.(b)
Solution:
Formula:
- PLAN: Probability $=\frac{\text { Number of favourable outcomes }}{\text { Number of total outcomes }}$
As, $x _1+x _2+x _3$ is odd.
So, all may be odd or one of them is odd and other two are even.
$\therefore$ Required probability
$ \frac{{ }^{2} C _1 \times{ }^{3} C _1 \times{ }^{4} C _1+{ }^{1} C _1 \times{ }^{2} C _1 \times{ }^{4} C _1+{ }^{2} C _1 \times{ }^{2} C _1 \times{ }^{3} C _1 { }^{1} C _1 \times{ }^{3} C _1 \times{ }^{3} C _1} {{ }^{3} C _1 \times{ }^{5} C _1 \times{ }^{7} C _1} $
$ =\frac{24+8+12+9}{105} $
$ =\frac{53}{105}$
BETA
