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# Probability with permutations and combinations

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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
+21017
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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
+24995
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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?

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

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

Apr 28, 2020