There are seven boys and five girls in a class. The teacher randomly selects three different students to answer questions. The f

Question

8 days ago

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There are seven boys and five girls in a class. The teacher randomly selects three different students to answer questions. The f

irst student is a girl, the second student is a boy, and the third student is a girl. Find the probability of this occuring.

Mathematics

2 answers:

Zina [3.9K] · Answer 1 · 2026-02-25T05:15:11+03:00

The probability of this scenario occurring is 7/66

Additional clarification

Probability is described as the chance of an event happening compared to the sample space.

$\large { \boxed {P(A) = \frac{\text{Number of Favorable Outcomes to A}}{\text {Total Number of Outcomes}} } }$

Permutation ( Arrangement )

Permutation signifies the various ways to arrange items.

$\large {\boxed {^nP_r = \frac{n!}{(n - r)!} } }$

Combination ( Selection )

Combination refers to the different ways to select items.

$\large {\boxed {^nC_r = \frac{n!}{r! (n - r)!} } }$

Let’s analyze the problem!

There are seven boys and five girls present in the class

yielding a total of 12 students.

The chance that the first selected is a girl

The likelihood of the first choice being a girl P(G₁):

$P(G_1) = \boxed {\frac{5}{12}}$

The odds of selecting a boy second P(B):

$P(B) = \boxed {\frac{7}{11}}$

The probability of picking a girl third P(G₂):

$P(G_2) = \boxed {\frac{4}{10}}$

The probability of the first selected being a girl, the second being a boy, and the third being a girl:

$\large {\boxed {P(G_1 \cap B \cap G_2) = \frac{5}{12} \times \frac{7}{11} \times \frac{4}{10} = \frac{7}{66} } }$

We can utilize permutations to derive this outcome.

From the 5 girls, arrange 2 for the first and third selections:

$^5P_2 = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 4 \times 5 = \boxed{20}$

From the 7 boys, arrange 1 for the second choice:

$^7P_1 = \frac{7!}{(7-1)!} = \frac{7!}{6!} = \boxed {7}$

The total count of arrangements for choosing 3 distinct students from a group of 12:

$^{12}P_3 = \frac{12!}{(12-3)!} = \frac{12!}{9!} = 10 \times 11 \times 12 = \boxed {1320}$

The probability of first selecting a girl, then a boy, and again a girl:

$\large {\boxed {P(G_1 \cap B \cap G_2) = \frac{20 \times 7}{1320} = \frac{7}{66} } }$

Explore more

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Answer specifics

Level: High School

Subject: Mathematics

Section: Probability

Keywords: Probability, Sample, Space, Six, Dice, Die, Binomial, Distribution, Mean, Variance, Standard Deviation

Svet_ta [4.3K] · Answer 2 · 2026-02-25T05:16:25+03:00

The likelihood of selecting one girl is calculated as $\frac{5}{12}$ . This is based on having 5 girls within a total of 12 students, and the probability of an event can be expressed as: $\frac{\text{# of things you want}}{\text{# of things are possible}}$ .

Using the same reasoning, for the next student, we have reduced the number of students by 1, leading to 11 possible outcomes instead of 12, giving us: $\frac{7}{11}$ , which represents the probability of selecting a boy as the second choice.

Lastly, the probability of choosing a girl for the third selection follows the same logic and is given as: $\frac{4}{10}$ .

However, we must combine these individual probabilities to determine the likelihood of this specific sequence of selections occurring:

$\frac{5}{12}*\frac{7}{11}*\frac{4}{10}=\frac{140}{1320}$

This simplifies to:

$\frac{7}{66}$