Answer to Question #105741 in Statistics and Probability for This website is good

Question

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

Answer 1

The number of possible 6-number combinations

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

The probability that you will match k numbers 

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

 

$$P(0)=\dfrac{\dbinom{6}{0}\dbinom{49-6}{6-0}}{\dbinom{49}{6}}=\dfrac{\dfrac{6!}{0!(6-0)!}\cdot\dfrac{43!}{6!(43-6)!}}{13983816}=$$ $$={1\cdot6096454\over 13983816}={6096454 \over 13983816}$$

$$P(1)=\dfrac{\dbinom{6}{1}\dbinom{49-6}{6-1}}{\dbinom{49}{6}}=\dfrac{\dfrac{6!}{1!(6-1)!}\cdot\dfrac{43!}{1!(43-1)!}}{13983816}=$$ $$={6\cdot962598\over 13983816}={5775588 \over 13983816}$$

$$P(2)=\dfrac{\dbinom{6}{2}\dbinom{49-6}{6-2}}{\dbinom{49}{6}}=\dfrac{\dfrac{6!}{2!(6-2)!}\cdot\dfrac{43!}{4!(43-4)!}}{13983816}=$$ $$={15\cdot123410\over 13983816}={1851150 \over 13983816}$$

$$P(3)=\dfrac{\dbinom{6}{3}\dbinom{49-6}{6-3}}{\dbinom{49}{6}}=\dfrac{\dfrac{6!}{3!(6-3)!}\cdot\dfrac{43!}{3!(6-3)!}}{b}=$$ $$={20\cdot12341\over 13983816}={246820 \over 13983816}$$

$$P(4)=\dfrac{\dbinom{6}{4}\dbinom{49-6}{6-4}}{\dbinom{49}{6}}=\dfrac{\dfrac{6!}{4!(6-4)!}\cdot\dfrac{43!}{2!(43-2)!}}{13983816}=$$ $$={15\cdot903\over 13983816}={13545 \over 13983816}$$

$$P(5)=\dfrac{\dbinom{6}{5}\dbinom{49-6}{6-5}}{\dbinom{49}{6}}=\dfrac{\dfrac{6!}{5!(6-5)!}\cdot\dfrac{43!}{1!(43-1)!}}{13983816}=$$ $$={6\cdot43\over 13983816}={258\over 13983816}$$

$$P(6)=\dfrac{\dbinom{6}{6}\dbinom{49-6}{6-6}}{\dbinom{49}{6}}=\dfrac{\dfrac{6!}{6!(6-6)!}\cdot\dfrac{43!}{0!(43-0)!}}{13983816}=$$ $$={1\cdot1\over 13983816}={1\over 13983816}$$

a. The probability that you will match 3 numbers  

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

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

$$P(\leq2)=P(0)+P(1)+P(2)=$$ $$={6096454 \over 13983816}+{5775588 \over 13983816}+{1851150 \over 13983816}={13723192\over 13983816}=$$ $$={245057 \over 249711}\approx0.98136$$

c. The expected number of matching number

$$E=0\cdot{6096454 \over 13983816}+1\cdot{5775588 \over 13983816}+2\cdot{1851150 \over 13983816}+$$

$$+3\cdot{246820 \over 13983816}+4\cdot{13545 \over 13983816}+5\cdot{258 \over 13983816}+$$

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