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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?

- $0 \\ $
- $\frac{325}{864} \\ $
- $\frac{525}{864} \\ $
- $\frac{5}{12}$

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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)**

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