Question 1092525: Find the probability that in tossing a fair coin 9 times, there will appear:
1. 6 tails
2. 5 heads
3. at least 2 tails
4. not more than 3 heads
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The general form of this kind of problem is nCr*p^r*(1-p)^(n-r). Here, p and 1-p are both a half, so the terms p^r*(1-p)^(n-r) are really (1/2)^9 or 1/518.
For 9C0, the probability of 0 heads (or 0 tails, if one wishes), it is 1
for 9C1, the probability of 1, it is 9!/1!*8!=9
and so forth.
6 tails is 9C6(1/2)^6((1/2)^3
84/512=0.1641
-----------
5 heads is 9C5(1/2)^5(1/2)^4=126/512=0.2461
--------------------
at least 2 tails is 1- p(1 tail)-p(0) tails)
p(1) is 9/512 and p(0) is 1/512
it is 502/512 or 0.9805
-------------------
not more than 3 heads is p(0)+p(1)+p(2)+p(3)
this is over a denominator of 512 (the 1/2^x*(1/2)^9-x
p(0) has a numerator of 1, p(1) has 9 p(2) has 9C2=36, and p(3) has 9C3 as a numerator, or 84
the probability is 130/512 or 0.2539
|
|
|
