Miscellaneous Question 57
In the given questions, two quantities are given. One as Quantity-I and the other is Quantity-II. You have to determine relationship between these two quantities and choose the appropriate options as given below :
(1) Quantity-I > II
(2) Quantity-I $<$ II
(3) Quantity-I $\leq$ II
(4) Quantity-I $\geq$ II
(5) Quantity-I = II (or) relationship cannot be established
- Consider the following probability cases.
Quantity-1: Four persons are chosen at random from a group of 3 men, 5 women and 4 children. What is the probability that exactly two of them are men ?
Quantity-2: A box contains 3 balloons of one shape, 4 balloons of another shape and 5 balloons of another shape. Three balloons of them are drawn at random, what is the probability that all the three are of different shapes?
Show Answer
Correct Answer: 57. (2)
Solution: 57. (2) Quantity I:
Total people
$=3+5+4=12 ; \rightarrow n(s)$
$={ }^{12} \mathrm{C}_{4}$
We select exactly two men out of 3 men and two persons from others
$\therefore n(\mathrm{E})={ }^{3} \mathrm{C} _{2} \times{ }^{9} \mathrm{C} _{2}$
Probability $=\frac{{ }^{3} \mathrm{C}_{2} \times{ }^{9} \mathrm{C} _{2}}{{ }^{12} \mathrm{C} _{4}}$
$=\frac{\frac{3 \times 9 \times 8}{1 \times 2}}{\frac{12 \times 11 \times 10 \times 9}{1 \times 2 \times 3 \times 4}}=\frac{12}{5}=0.22$
Quantity II:
Total $=3+4+5=12$
$n(\mathrm{S})={ }^{12} \mathrm{C}_{3}=\frac{12 \times 11 \times 10}{1 \times 2 \times 3}$
$=220$
$n(\mathrm{E})={ }^{3} \mathrm{C} _{1} \times{ }^{4} \mathrm{C} _{1} \times{ }^{5} \mathrm{C} _{1}$
$p=\frac{60}{220}=\frac{3}{11}=0.27$
$\therefore \quad $ Quantity-I $<$ II
BETA
