There are seven boys and five girls in a class. The teacher randomly selects three different students to answer questions. The first student is a girl, the second student is a boy, and the third student is a girl. Find the probability of this occuring.

Respuesta :

The probability of this occuring is 7/66

Further explanation

The probability of an event is defined as the possibility of an event occurring against sample space.

[tex]\large { \boxed {P(A) = \frac{\text{Number of Favorable Outcomes to A}}{\text {Total Number of Outcomes}} } }[/tex]

Permutation ( Arrangement )

Permutation is the number of ways to arrange objects.

[tex]\large {\boxed {^nP_r = \frac{n!}{(n - r)!} } }[/tex]

Combination ( Selection )

Combination is the number of ways to select objects.

[tex]\large {\boxed {^nC_r = \frac{n!}{r! (n - r)!} } }[/tex]

Let us tackle the problem!

There are seven boys and five girls in a class

There are a total of 12 students.

The possibility of being elected first is a woman

The probability of being choosen first is a girl P(G₁):

[tex]P(G_1) = \boxed {\frac{5}{12}}[/tex]

The probability of being choosen second is a boy P(B):

[tex]P(B) = \boxed {\frac{7}{11}}[/tex]

The probability of being choosen third is a girl P(G₂):

[tex]P(G_2) = \boxed {\frac{4}{10}}[/tex]

The probability of choosing the first student is a girl, the second student is a boy, and the third student is a girl:

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

We can also use permutations to get this answer.

From 5 girls we will arrange 2 girls as first and third student:

[tex]^5P_2 = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 4 \times 5 = \boxed{20}[/tex]

From 7 boys we will arrange 1 boys as second student:

[tex]^7P_1 = \frac{7!}{(7-1)!} = \frac{7!}{6!} = \boxed {7}[/tex]

The total number of arrangement 3 different students from total of 12 students.

[tex]^{12}P_3 = \frac{12!}{(12-3)!} = \frac{12!}{9!} = 10 \times 11 \times 12 = \boxed {1320}[/tex]

The probability of choosing the first student is a girl, the second student is a boy, and the third student is a girl:

[tex]\large {\boxed {P(G_1 \cap B \cap G_2) = \frac{20 \times 7}{1320} = \frac{7}{66} } }[/tex]

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Answer details

Grade: High School

Subject: Mathematics

Chapter: Probability

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

Ver imagen johanrusli

The probability the first student is a girl, the second student is a boy, and the third student is a girl is 0.1061

The given parameters are:

[tex]B = 7[/tex] --- boys

[tex]G = 5[/tex] --- girls

So, the total number of students is 12

When the first student (a girl) is selected, the probability is:

[tex]Pr = \frac{G}{Total}[/tex]

[tex]Pr = \frac{5}{12}[/tex]

Now, there are 11 students; because the selection is without replacement

When the second student (a boy) is selected, the probability is:

[tex]Pr = \frac{B}{Total - 1}[/tex]

[tex]Pr = \frac{7}{11}[/tex]

Now, there are 10 students;

When the third student (a girl) is selected, the probability is:

[tex]Pr = \frac{G-1}{Total-2}[/tex]

[tex]Pr = \frac{4}{10}[/tex]

So, the probability of the selection is:

[tex]P = \frac{5}{12} \times \frac{7}{11} \times \frac{4}{10}[/tex]

[tex]P = 0.1061[/tex]

Hence, the probability is 0.1061

Read more about probabilities at:

https://brainly.com/question/8652467