Question

a family has 8 girls and 4 boys. a total of 3 children must be chosen to speak on behalf of their family at a local benefit. what is the probability that no girls and 3 boys will be chosen

Answer 1

Answer:  The required probability that no girls and 3 boys will be chosen is $\dfrac{1}{55}.$

Step-by-step explanation:  Given that a family has 8 girls and 4 boys. a total of 3 children must be chosen to speak on behalf of their family at a local benefit.

We are to find the probability that no girls and 3 boys will be chosen.

Let, S denotes the sample space for the experiment of choosing 3 children and E be the event that no girls and 3 boys will be chosen.

Then, we have

$n(S)=^{8+4}C_3=^{12}C^3=\dfrac{12!}{3!(12-3)!}=\dfrac{12\times11\times10\times9!}{3\times2\times1\times9!}=4\times5\times11=220,\\\\\\n(E)=^4C_3=\dfrac{4!}{3!(4-3)!}=\dfrac{4\times3!}{3!\times1}=4.$

Therefore, the probability of event E will be

$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{4}{220}=\dfrac{1}{55}.$

Thus, the required probability that no girls and 3 boys will be chosen is $\dfrac{1}{55}.$

Answer 2

Answer
1/55

Explanation
The probability of an event is a chance of it happening. It is calculated as;
Probability =(fouvarable outcome)/(total outcome)
Since we are finding the probability of no girl will be chosen, it is like finding the probability of choosing 3 boys.
The probability of choosing the first boy =4/12
The probability of choosing the second boy =3/11
The probability of choosing the third boy =2/10
The probability of choosing the 3 boys will be,
=4/12×3/11×2/10
=24/1320
=1/55