Question

An experiment consists of four tosses of a coin. Denoting the outcomes: HHTH,THTT,... and assuming that all 16 outcomes are equally likely, find the probability distribution for the total number of heads.

Answer 1

The possible outcomes for heads is HHTH, THTT,HHTT, and HHHH.

Answer 2

Therefore the probability distribution for ther total number of heads when X= 0,1,2,3 is 3+1+2+4=10 total number of heads.

Answer 3

distribution means if you put \(X\) as the total number of head, you have to compute the following
\[P(X=0),P(X=1),P(X=2),P(X)=3,P(X=4)\]

Answer 4

these are respectively
\[\frac{1}{2^4}\]
\[\frac{4}{2^4}\]
\[\frac{6}{2^4}\]
\[\frac{4}{2^4}\]
\[\frac{1}{2^4}\]

Answer 5

the denominator is 16, the number of total possible outcomes
the numerator is the number of ways to pick 0, 1, 2, 3, 4 items out of 4 respectively

Answer 6

So I see that you got 2 from the selection of "heads" or "tails" and ^4 from their being four outcomes in each individual toss. What I'm confused about are the numerators "1,4,6,4, and 1".

Answer 7

pascals triangle is one place

Answer 8

1 4 6 4 1
first number is the number of ways to choose 0 out of 4 (no tails)
second is the number of ways to choose 1 out of 4 (one tail)
here they are \((h, h, h, t), (h, h, t, h), (h, t, h, h), (t, h, h, h)\)

Answer 9

third number is the number of ways to choose 2 out of 4
fourth number is the number of ways to choose 3 out of 4 (same as choosine 1) and last number is the number of ways to choose 4 out of 4

Answer 10

Oh okay. I get it now, I wasn't understanding the concept properly but now I do. Thank you.