GATE Electrical 2018 | GA | Question: 8

Question

A class of twelve children has two more boys than girls. A group of three children are randomly picked from this class to accompany the teacher on a field trip. What is the probability that the group accompanying the teacher contains more girls than boys?

• A. $0 \\ $
• B. $\frac{325}{864} \\ $
• C. $\frac{525}{864} \\ $
• D. $\frac{5}{12}$

Answer 1

In the class, $ Boys = Girls + 2 $

& $ Boys + Girls = 12 $

$\color{blue}{\text{∴ Boys = 7 & Girls = 5}}$

Now, We have to choose $3$ students from $12$ students which can be done in ${^{12}C_3 }$ 

But among these $3$ students, girls should be more than the boys

It can be done in $2$ ways

either in the field trip, there are $2$ girls and $1$ boys OR there are $3$ girls

∴  Choosing $2$ girls from $5$ girls AND choosing $1$ boys from $7$ boys = ${^5C_2} \times {^7C_1}$

    Choosing $3$ girls from $5$ girls =  ${^5C_3}$

$\color{maroon}{\text{∴ Probability that the group accompanying the teacher contains more girls than boys }}$
$=\color{purple}{\dfrac{\text{number of favorable ways}}{\text{total number of ways}}}$

$=\dfrac{{^5C_2 } \times {^7C_1} + {^5C_3 }}{{^{12}C_3}}$

$=\dfrac{(10)\times(7)+(10)}{220}$

$=\dfrac{80}{220}=\dfrac{4}{11}$

$\color{green}{\text{Hence, the answer is}}\color{purple} \ {\text{None of these}}$

This was marks to all in GATE

Comments

This was marks to all because no option was matching with(4/11)...which is approx .0.36...

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How could u include the case of selecting all girls when it is clearly mentioned in the questions that girls are more than boys (so in this sense it is implied at least one boy has to be there,otherwise how could u compare against girls ?? ))

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@s_dr_13

$3$ Girls and $0$ Boys is a valid case.

Read question carefully again. 

Answer 2

Let ,total boys = x and total girls = y .  So according to given condition , x + y = 12 and x = y + 2 ... So we get x = 7 and y = 5  ie 7 boys and 5 girls.

Now , Since we have to pick 3 children in such a way that Girls should be more than boys in the group. So , either 3 Girls and 0 boy (or)  2 Girls and 1 boy will be in group.

So , Required Probability = [C(5,3)*C(7,0) + C(5,2)*C(7,1)]  / C(12,3)

Answer 3

In the class, Boys = Girls + 2

& Boys + Girls = 12

∴ Boys = 7 & Girls = 5

Now, We have to choose 3 students from 12 students which can be done in  ${^{12}C_3}$

But among these 3 students, girls should be more than the boys

It can be done in 2 ways

either in the field trip, there are 2 girls and 1 boys OR there are 3 girls

∴  Choosing 2 girls from 5 girls AND choosing 1 boys from 7 boys = ${^{5}C_2}$ * ${^{7}C_1}$

    Choosing 3 girls from 5 girls =  ${^{5}C_3}$

∴ Probability that the group accompanying the teacher contains more girls than boys = $\dfrac{number of favorable ways}{total number of ways}$

= $\dfrac{{5 \choose 2} * {7 \choose 1} + {5 \choose 3}}{{12 \choose 3}}$

= $\dfrac{(10)*(7)+(10)}{220}$

= $\dfrac{80}{220}$

= 0.36 ≅ $\dfrac{325}{864}$

Hence, the answer is option B)