# Thread: probability distribution

1. ## probability distribution

how would i prepare a probability distribution of the X variable for this question?

You have a biased coin or unfair coin, such that, on each toss, the probability of observing a head is 0.25 and the probability of observing a tail is 0.75. Let X be the variable = number of heads observed in 4 tosses.

2. Originally Posted by skhan
how would i prepare a probability distribution of the X variable for this question?

You have a biased coin or unfair coin, such that, on each toss, the probability of observing a head is 0.25 and the probability of observing a tail is 0.75. Let X be the variable = number of heads observed in 4 tosses.
The possible outcomes are:

hhhh p=0.25^4

thhh p=0.25^3 0.75
hthh p=0.25^3 0.75
hhth p=0.25^3 0.75
hhht p=0.25^3 0.75

tthh p=0.25^2 0.75^2
thth p=0.25^2 0.75^2
thht p=0.25^2 0.75^2
htth p=0.25^2 0.75^2
htht p=0.25^2 0.75^2
hhtt p=0.25^2 0.75^2

ttth p=0.25 0.75^3
ttht p=0.25 0.75^3
thtt p=0.25 0.75^3
httt p=0.25 0.75^3

tttt p=0.75^4

So

P(1 heads)=4 p(httt)=4 0.25 0.75^3
P(2 heads)=6 p(hhtt)=6 0.25^2 0.75^2
P(3 heads)=4 p(hhht)=4 0.25^3 0.75

This can also be done using the formulae for the binomial distribution.

RonL

3. thanks Ron

from the probabilities that you gave me, i tried to compute the average however i am getting an answer of avg = 1.14 whereas the right answer is given as avg = 1.0

can you spot where i am going wrong?

P(0 heads)=p(tttt)=0.75^4 = 0.31640
P(1 heads)=4 p(httt)=4 0.25 0.75^3 = 0.4218
P(2 heads)=6 p(hhtt)=6 0.25^2 0.75^2 = 0.21093
P(3 heads)=4 p(hhht)=6 0.25^3 0.75 = 0.07031
P(4 heads)=p(hhhh)=6 0.25^4 = 0.0234375

avg = 0(.31640) + 1(0.4218) + 2(0.21093) + 3(0.07031) + 4(0.02343)
= 0.4218 + 0.42186 + 0.21093 + 0.09372
= 1.14

4. Originally Posted by skhan
thanks Ron

from the probabilities that you gave me, i tried to compute the average however i am getting an answer of avg = 1.14 whereas the right answer is given as avg = 1.0

can you spot where i am going wrong?

P(0 heads)=p(tttt)=0.75^4 = 0.31640
P(1 heads)=4 p(httt)=4 0.25 0.75^3 = 0.4218
P(2 heads)=6 p(hhtt)=6 0.25^2 0.75^2 = 0.21093
P(3 heads)=4 p(hhht)=6 0.25^3 0.75 = 0.07031
P(4 heads)=p(hhhh)=6 0.25^4 = 0.0234375

avg = 0(.31640) + 1(0.4218) + 2(0.21093) + 3(0.07031) + 4(0.02343)
= 0.4218 + 0.42186 + 0.21093 + 0.09372
= 1.14
I'm afraid that is because some typos crept in during the cut and past job
used to compite the table, it should read: