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# Probabilities win a pick 6 game where six numbers are draw from the set 1 through 49 and get 24 heads in a row when you flip a fair coin,

probabilities independent and dependent events

### 2 Answers by Expert Tutors

Arthur D. | Effective Mathematics TutorEffective Mathematics Tutor
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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:
TT(tails then tails again)
See ! 1 out of 4 possibilities
probability of three heads in a row:
here are all the possibilities:
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 !

Joshua S. | An Astrophysicist Who Teaches Just About AnythingAn Astrophysicist Who Teaches Just About...
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Probabilities are all about counting your options. There are formulae you can memorize to answer either of these questions (associated with combinations without replacement and binomial probabilities), but there are simpler ways of doing this as well.

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.