probabilities independent and dependent events

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picking 6 numbers from a set of 49 numbers is a combination problem

use the formula (n!)/(r!)(n-r)! where n=49 and r=6

! is read factorial and means, for example, 5!=5*4*3*2*1, 8!=8*7*6*5*4*3*2*1

(49!)/(6!)(49-6)!=(49!)/(6!)(43!)

before we continue observe the following: (8*7*6*5*4*3*2*1)/(4*3*2*1)=8*7*6*5 because the

two (4*3*2*1)'s cancel each other out; one is in the numerator and one is in the denominator

continuing on:

(49*48*47*46*45*44*43*42*41*40...)/(6*5*4*3*2*1)(43*42*41*40...)=

(49*48*47*46*45*44)/(6*5*4*3*2*1)=

let's stop a minute; the two(43*42*41*40*...*4*3*2*1)'s cancel each other out

(49*48*47*46*45*44)/(6*5*4*3*2*1) can be simplified

48/6=8, 45/(5*3)=3, 46/2=23, and 44/4=11

now we have after simplifying:(49*8*47*23*3*11)=13,983,816

if you did not simplify you would get 10,068,347,520/720=13,983,816

the probability of getting 24 heads in arrow is (1/2)^24

the probability of getting heads on one toss is 1/2(you have one head and one tail for two possibilities)

the probability of getting two heads on two tosses is (1/2)(1/2)=1/4

the probability of getting three heads in arrow is (1/2)(1/2)(1/2)=(1/2)^3=1/8

four heads in a row is (1/2)^4=1/16

five heads in arrow is (1/2)^5=1/32

twenty-four heads in a row is (1/2)^24=1/16,777,216

a little something extra:

probability of two heads in a row:

here are all the possibilities:

HH(heads and heads)

HT(heads and tails)

TH(tails then heads)

TT(tails then tails again)

See ! 1 out of 4 possibilities

probability of three heads in a row:

here are all the possibilities:

HHH(heads, heads, heads)

HHT(heads, then heads again, then tails)

THH(tails first, then heads, and then heads again)

HTH(heads, then tails, then heads again) Do you see the picture ?

TTT

TTH

HTT

THT

the probability of three heads in a row is one out of eight possibilities

probability= 1/8

Looking at this gives you some insight into why we have formulas to help us

to solve problems. Imagine trying to do this for 24 heads !!

The same applies for the first problem; picking 6 numbers from forty-nine numbers. We used a formula !

For the first problem, you would typically use the *combination* 49 choose 6 as the starting point (aka the denominator for the probability of hitting the jackpot). To understand the details why this is so, check out the Wikipedia page that documents this very problem! https://en.wikipedia.org/wiki/Lottery_mathematics

For the second problem, you are simply asking for a binomial probability of 24 independent coin flips. You can build up your intuition for this by asking yourself, "What's the probability of getting 1 head in a row? 2 heads in a row? 3? etc." You'll see a pattern of a p^n showing up which should lead you to the correct answer.

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