Question

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.

Answer 1

The probability of picking one girl would be $\frac{5}{12}$. That is because there are 5 girls out of the 12 students, and the probability of an event occuring is: $\frac{\text{# of things you want}}{\text{# of things are possible}}$.

Using that same logic, the next student should be easier. We reduced the student population by 1, so we have 11 possible ways it can happen now instead of 12, so that gives us:$\frac{7}{11}$, for the probability of picking a boy as the second pick.

And lastly, using the same logic shown above, the probability of picking a girl on the third pick would be:$\frac{4}{10}$.

We are not done, though. We have the separate probabilities, but now we have to multiply then together to figure out the probability of this exact event happening:

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

Which when reduced is:

$\frac{7}{66}$

Answer 2

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.

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

Permutation ( Arrangement )

Permutation is the number of ways to arrange objects.

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

Combination ( Selection )

Combination is the number of ways to select objects.

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

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):

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

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

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

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

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

The probability of choosing the first student is a girl, the second student is a boy, and the third student is 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 also use permutations to get this answer.

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

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

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

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

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

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

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

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

Learn more

• Different Birthdays : brainly.com/question/7567074
• Dependent or Independent Events : brainly.com/question/12029535
• Mutually exclusive : brainly.com/question/3464581

Answer details

Grade: High School

Subject: Mathematics

Chapter: Probability

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