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]
Learn more
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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
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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