Question #105741
In Lotto 649, you must select 6 numbers from 1-49. Calculate the following :

a. The probability that you will match 3 number
b. The probability that you will match 2 or less number
c. The expected number of matching number
1
Expert's answer
2020-03-17T14:48:29-0400

The number of possible 6-number combinations


49C6=49!6!(496)!=13983816_{49}C_{6}={49! \over 6!(49-6)!}=13983816

The probability that you will match k numbers 


P(k)=(6k)(4966k)(496), k=0,1,2,3,4,5,6P(k)=\dfrac{\dbinom{6}{k}\dbinom{49-6}{6-k}}{\dbinom{49}{6}},\ k=0,1,2,3,4,5,6

 

P(0)=(60)(49660)(496)=6!0!(60)!43!6!(436)!13983816=P(0)=\dfrac{\dbinom{6}{0}\dbinom{49-6}{6-0}}{\dbinom{49}{6}}=\dfrac{\dfrac{6!}{0!(6-0)!}\cdot\dfrac{43!}{6!(43-6)!}}{13983816}==1609645413983816=609645413983816={1\cdot6096454\over 13983816}={6096454 \over 13983816}

P(1)=(61)(49661)(496)=6!1!(61)!43!1!(431)!13983816=P(1)=\dfrac{\dbinom{6}{1}\dbinom{49-6}{6-1}}{\dbinom{49}{6}}=\dfrac{\dfrac{6!}{1!(6-1)!}\cdot\dfrac{43!}{1!(43-1)!}}{13983816}==696259813983816=577558813983816={6\cdot962598\over 13983816}={5775588 \over 13983816}

P(2)=(62)(49662)(496)=6!2!(62)!43!4!(434)!13983816=P(2)=\dfrac{\dbinom{6}{2}\dbinom{49-6}{6-2}}{\dbinom{49}{6}}=\dfrac{\dfrac{6!}{2!(6-2)!}\cdot\dfrac{43!}{4!(43-4)!}}{13983816}==1512341013983816=185115013983816={15\cdot123410\over 13983816}={1851150 \over 13983816}


P(3)=(63)(49663)(496)=6!3!(63)!43!3!(63)!b=P(3)=\dfrac{\dbinom{6}{3}\dbinom{49-6}{6-3}}{\dbinom{49}{6}}=\dfrac{\dfrac{6!}{3!(6-3)!}\cdot\dfrac{43!}{3!(6-3)!}}{b}==201234113983816=24682013983816={20\cdot12341\over 13983816}={246820 \over 13983816}

P(4)=(64)(49664)(496)=6!4!(64)!43!2!(432)!13983816=P(4)=\dfrac{\dbinom{6}{4}\dbinom{49-6}{6-4}}{\dbinom{49}{6}}=\dfrac{\dfrac{6!}{4!(6-4)!}\cdot\dfrac{43!}{2!(43-2)!}}{13983816}==1590313983816=1354513983816={15\cdot903\over 13983816}={13545 \over 13983816}


P(5)=(65)(49665)(496)=6!5!(65)!43!1!(431)!13983816=P(5)=\dfrac{\dbinom{6}{5}\dbinom{49-6}{6-5}}{\dbinom{49}{6}}=\dfrac{\dfrac{6!}{5!(6-5)!}\cdot\dfrac{43!}{1!(43-1)!}}{13983816}==64313983816=25813983816={6\cdot43\over 13983816}={258\over 13983816}

P(6)=(66)(49666)(496)=6!6!(66)!43!0!(430)!13983816=P(6)=\dfrac{\dbinom{6}{6}\dbinom{49-6}{6-6}}{\dbinom{49}{6}}=\dfrac{\dfrac{6!}{6!(6-6)!}\cdot\dfrac{43!}{0!(43-0)!}}{13983816}==1113983816=113983816={1\cdot1\over 13983816}={1\over 13983816}

a. The probability that you will match 3 numbers  


P(3)=24682013983816=88154994220.01765P(3)={246820 \over 13983816}={8815 \over 499422}\approx0.01765

b. The probability that you will match 2 or less numbers 


P(2)=P(0)+P(1)+P(2)=P(\leq2)=P(0)+P(1)+P(2)==609645413983816+577558813983816+185115013983816=1372319213983816=={6096454 \over 13983816}+{5775588 \over 13983816}+{1851150 \over 13983816}={13723192\over 13983816}==2450572497110.98136={245057 \over 249711}\approx0.98136

c. The expected number of matching number


E=0609645413983816+1577558813983816+2185115013983816+E=0\cdot{6096454 \over 13983816}+1\cdot{5775588 \over 13983816}+2\cdot{1851150 \over 13983816}+

+324682013983816+41354513983816+525813983816++3\cdot{246820 \over 13983816}+4\cdot{13545 \over 13983816}+5\cdot{258 \over 13983816}+

+6113983816=1027382413983816=36490.7347+6\cdot{1 \over 13983816}={10273824 \over 13983816}={36 \over 49}\approx0.7347


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