Math post: Probability

Here is a long needed mathematics blog post. This weeks subject is Probability and the question is as follows:

Suppose that a biased coin that lands on heads with probability p is flipped 10 times. Given that a total number of 6 heads result, find the conditional probability that the first 3 outcomes are

(a) h,t,t (meaning that the first flip is a head, the second is a tail, and the third is tails)

(b) t,h,t

Answers:

(a) We will have to know one formula to calculate these probabilities, as well as some basic knowledge on summations and the choose function. The needed formula is P(A|B)= P(AB)/P(B). This reads as the probability of an event A given B is equal to the probability of A and B both occurring over the probability of B occuring.

Thus for (a), we are trying to find P(H,T,T|6 Heads). We have:

P(H,T,T|6 Heads)= P(H,T,T and 6 heads)/ P(6 heads)

=P(H,T,T)*P(6 heads |H,T,T)/ P(6 heads)

since P(6 heads|H,T,T)= P( H,T,T and 6 heads)/ P(H,T,T) implies that P( H,T,T and 6 heads)=P(6 heads|H,T,T)* P(H,T,T)

= (p*(1-p)*(1-p))* (7 choose 5)*p^5*(1-p)^2) / (10 choose 6)*p^6*(1-p)^4

The logic in this is first give the probability of H,T,T. That is give the probability of flipping heads (p) times the probability of flipping tails twice (1-p)(1-p). (from now on I will write q for 1-p) Thus we have P(H,T,T). Then we must solve for P(6 heads |H,T,T). Think about how many ways the other five heads can be given if we have already been give one. We have already done three of the ten flips, so we have seven more and we need to get exactly five heads in those flips. Thus we get the (7 choose 5). We then need the probability of getting exactly 5 heads and 2 tails: p^5*p^2. Thus we have the probability of P(6 heads |H,T,T). For the probability of 6 heads we use the same logic as before but now no flips have been used, so we have ten flips to land 6 heads. There are (10 choose 6) ways of doing this. We then consider the probability of this occurring. This is p^6*q^4.

We thus receive the equation:

=(p(q)(q)(7 choose 5)(p^5)(q^2))/ ((10 choose 6)(p^6)(q^4))

=((p^6)(q^4)(7 choose 5))/ ((p^6)(q^4)(10 choose 6)) by combining the p's and q's (remember (x^a)(x^b)=(x^(a+b)))

=(7 choose 5)/ (10 choose 6)

=21/210 (calculator or (n choose i)= (n!/ (n-i)!(i!)) where n!= n(n-1)(n-2)...*2*1 ¹)

=1/10

(b) P(T,H,T| 6 heads)= P(T,H,T and 6 heads)/ P(6 heads)

=P(T,H,T)*P(6 heads| T,H,T)/ (P(6 heads))

=((q^2)(p)(7 choose 5)(p^5)(q^2))/ ((10 choose 7)(p^6)(q^4))

=((p^6)(q^4)(7 choose 5))/ ((p^6)(q^4)(10 choose 6))

=(7 choose 5)/ (10 choose 6)

=21/210

=1/10

¹ (7 choose 5) = 7!/ (2!)(5!)= (7*6*5*4*3*2*1)/ (2*1)( 5*4*3*2*1)= 7*6/ 2= 42/2=21

(10 choose 6)= 10!/ (4!)(6!)= (10*9*8*7)/ (4*3*2*1)= 5040/ 24=210

I hope that you enjoyed this. The next post will be a cultural one. I hope to be posting math based posts at a higher rate in the future, so for those who are interested "Yay!", for others... sorry, but I hope it makes you feel better that I am able to understand the material better if I work to break it down for you, so please let me know if you need any thing else explained in more detail.