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There are 10 students in a class: 6 boys and 4 girls. If the teacher picks a group of 3 at random, what is the probability that everyone in the group is a girl?

 Apr 27, 2020
 #1
avatar+21017 
+1

First person is a girl:       4/10

Second person is a girl:  3/9                       (3 girls remain; 9 students remain)

Third person is a girl:      2/8                       (2 girls remain; 8 students remain)

 

For all three to be girls:  (4/10) x (3/9) x (2/8)  =  1/30

 Apr 27, 2020
 #2
avatar+24995 
+2

There are 10 students in a class: 6 boys and 4 girls. If the teacher picks a group of 3 at random,

what is the probability that everyone in the group is a girl?

 

Hypergeometric distribution:
\(\text{$N$ is the number of $\mathbf{students} = 10$}, \\ \text{$b$ is the number of $\mathbf{boys} = 6$}, \\ \text{$g$ is the number of $\mathbf{girls}= 4$}, \\ \text{$k$ the teacher picks a $\mathbf{boy} = 0 $}, \\ \text{$n$ the teacher picks a $\mathbf{girl} = 3 $}.\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{ \dbinom{b}{k} \dbinom{ g }{n} }{ \dbinom{N}{n+k} }} \\\\ &=& \dfrac{ \dbinom{6}{0}_{\text{boys}} \dbinom{4}{3}_{\text{girls}} } { \dbinom{10}{3}_{\text{all}} } \\\\ && \boxed{\dbinom{6}{0} = 1,\\ \dbinom{4}{3}=\binom{4}{4-3}=\binom{4}{1}=4,\\ \dbinom{10}{3}=\dfrac{10}{3}\cdot \dfrac{9}{2}\cdot \dfrac{8}{1} = 5\cdot 3 \cdot 8 } \\\\ &=& \dfrac{ 1\cdot 4 } { 5\cdot 3 \cdot 8 } \\\\ &=& \dfrac{ 1 } { 5\cdot 3 \cdot 2 } \\\\ &=&\mathbf{ \dfrac{ 1 } { 30 }} \\\\ &=& 0.03333333333\quad (=3.3\ \%) \\ \hline \end{array}\)

 

Source see: https://en.wikipedia.org/wiki/Hypergeometric_distribution

 

laugh

 Apr 28, 2020

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