Statistics: Binomial Distribution?

Question

Problem: A fair coin is tossed. How likely is it to land half heads, half tails if it is tossed:
a.) 2 times; 4 times; 10 times; 20 times

Answer is:
n=2: 50%
n=4: 37.5%
n=10: 24.6%
n=20: 17.6%

the question is how to get those answers? i cant seem to find any similar examples in my book. thanks.

Answer 1

Use the equation for solving binomial distributions which is described in detail here.

http://mathworld.wolfram.com/BinomialDistribution.html

You take the binomial coefficient times the probability of each desired possibility to the power of desired successes. The binomial coefficient is equal to n!/[(n-k)!k!] or BC(n,k). The actual way to write out BC can be seen here

http://mathworld.wolfram.com/BinomialCoefficient.html

,however there is no easy way to write it out on Yahoo! Answers so I use this convention. The equation for solving a binomial distribution is

BC(n,k) * P^k * (1-P)^(n-k)

I'll explain how to go about solving your problem.

n equals the number of tosses
k equals the number of desired successful tosses for P.
--> Because you want the number of tosses to be evenly
--> distributed, k will equal n/2. If you wanted the number
--> of tosses to favor P, you would increase P by an
--> integer N.
P equals the probability of landing on heads.
--> The probability of landing on heads is 1/2.
p equals the probability of landing on something else, namely tails.
--> The probability of landings on tails is 1/2.
--> Notice that p = 1 - P

eg.
2 tosses with an equal number of possibilities
BC( 2,1) * (1/2)^1 * (1/2)^1 = 2 * (1/2) * (1/2) = 1/2 = 50%

4 tosses with an equal number of possibilities
BC(4,2) * (1/2)^2 * (1/2)^2 = 6 * (1/4) * (1/4) = 6/16 = 37.5%

BC(10,5) * (1/2)^5 * (1/2)^5 = 252 * (1/32) * (1/32) =
--> 252/1024 = 24.6%

=================== ========================

Let's say P were heads and we wanted to favor it 6 to 4 over tails. The equation would then be

Pp(10,6)

BC(10,6) * (1/2)^6 * (1/2)^4 =
--> 210 * (1/64) * (1/16) = 210/1024 = 20.5%

The probability of landing on heads 6 times over 10 tosses
is 20.5%.

=================== ========================

If you wanted to determine the probability of rolling a one on a six-sided dice N times out of R rolls, this is how you would do it.

P = 1/6
p = 1 - P = 5/6

BC( R, N) * (1/6)^N * (5/6)*(R-N)

This should hopefully make it easy for you to understand how to use this equation.

Answer 2

Is there a table of binomial distribution in your book/notes?
You can also do probabilities by hand but it's time consuming. It requires that you know the total number of possible outcomes. For n = 2 this is easy:
TT
HH
TH
HT
So in 2/4 times you have the event you're interested in = .5 probability (i.e.50%).

For n=4 there's 16 (2^4)possible outcomes and there's 6 possible events that suit your outcome
TTHH
HHTT
HTHT
THTH
THHT
HTTH
That's 6/16 = .375
Obviously for n=10 this increases a lot. So use a table like below (if this is homework and they want it by hand then don't use the table - do calculations like above).


To read a binomial table: you need to know n. An n=10 (left hand side) means 10 coin tosses in your case. Then reading down the page the numbers r 0-10 will mean the probability that you have e.g. a head. But you want half heads and half tails which means you read off the values for 5 (i.e. the probability of 5 heads from 10 tosses of the coin) which is .246 (24.6%). THe right hand side table is the same but for 20 tosses of a coin. Read it the same way. You want 10 which gives you 17.6

I CANT GET THE TABLE TO APPEAR PROPERLY - so the left most column starts with r as the heading and reads 0-10 down the page then has 2 numbers next to each number. Then the Column for 0-20 is meant to be the 4th column along to the right. The problem is it prints below the table for n=10 so basically shift (in your mind) 11-20 to the far right so it lines up with the 2nd lot of numbers which starts 0-10. Hope that made sense.

Table 1: Selected Binomial Distributions
p = .5

n = 10 n = 20
rindcom rindcom
0.0011- 00+1-
1.010.999 10+1-
2.044.989 20+1-
3.117.945 3.0011-
4.205.828 4.005.999
5.246.623 5.015.994
6.205.377 6.037.979
7.117.172 7.074.942
8.044.055 8.120.868
9.010.011 9.160.748
10.001.001 10.176.588
11.160.412
12.120.252
13.074.132
14.037.058
15.015.021
16.005.006
17.001.001
180+0+
190+0+
200+0+

Answer 3

n!=1x2x3..n

answer = n!/((n/2)!*(n/2)!*(2^n))

n=4

n!=1x2x3x4=24
(n/2)!=1x2=2
2^n=2x2x2x2=16

answer=24/(2*2*16) = 0.375

enjoy:)