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3 years ago

Paul has a standard deck of cards. what is the probability he will choose a 2?

Mathematics

1 answer:

Zigmanuir [339]3 years ago

6 0

$|\Omega|=52\\
|A|=4\\\\
P(A)=\dfrac{4}{52}=\dfrac{1}{13}\approx7,7\%$

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Kobotan [32]

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See explanation

Step-by-step explanation:

There are three possible cases:

1. Point N lies between M and P, then MN + NP = MP. Consider needed difference:

$\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{NP}-\dfrac{MN}{MN+NP}=\dfrac{MN(MN+NP)-MN\cdot NP}{NP(MN+NP)}=\\ \\=\dfrac{MN^2+MN\cdot NP-MN\cdot NP}{NP(MN+NP)}=\dfrac{MN^2}{NP(MN+NP)}$

2. Point N lies to the right from point P, then MP + PN = MN. Consider needed difference:

$\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MP+PN}{NP}-\dfrac{MP+PN}{MP}=\dfrac{MP}{NP}+1-1-\dfrac{NP}{MP}=\dfrac{MP^2-NP^2}{NP\cdot MP}$

3. Point N lies to the left from point M, then NM + MP = NP. Consider needed difference:

$\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{MN+MP}-\dfrac{MN}{MP}=\dfrac{MN\cdot MP-MN(MN+MP)}{MP(MN+MP)}=\\ \\=\dfrac{MN\cdot MP-MN^2-MN\cdot MP}{MP(MN+MP)}=\dfrac{-MN^2}{MP(MN+MP)}$

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