probability (1 Viewer)

stressedadfff · Aug 15, 2021

hey guys anyone know 12d?

Screen Shot 2021-08-15 at 1.15.07 pm.png

CM_Tutor · Aug 15, 2021

Let B = Beagle chosen, G = German Shepherd chosen, and L = Labrador chosen.

$\begin{align*} P(\text{one Beagle and one German Shepherd}) &= P(\text{B and G in any order}) \\ &= P(\text{BG or GB}) \\ &= P(\text{BG}) + P(\text{GB}) \\ &= \cfrac{5}{15} \times \cfrac{6}{14} + \cfrac{6}{15} \times \cfrac{5}{14} \\ &= \cfrac{1}{3} \times \cfrac{3}{7} + \cfrac{2}{5} \times \cfrac{5}{14} \\ &= \cfrac{1}{7} + \cfrac{2}{14} \\ &= 2 \times \cfrac{1}{7} \\ &= \cfrac{2}{7} \end{align*}$

fnsndgfg · Aug 17, 2021

An alternate way to CM_Tutor is by using combinatorics, where the probability is:
(amount of combinations with only beagle and german shepherd)/(Amount of possible combinations)
(5C1 x 6C1)/(15C2)
=30/105
=2/7

probability (1 Viewer)

stressedadfff

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CM_Tutor

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fnsndgfg

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